title stringlengths 1 100 | titleSlug stringlengths 3 77 | Java int64 0 1 | Python3 int64 1 1 | content stringlengths 28 44.4k | voteCount int64 0 3.67k | question_content stringlengths 65 5k | question_hints stringclasses 970
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Problems with Python3 and Multiple Solutions | merge-k-sorted-lists | 0 | 1 | **Python 2 Solution:**\n```\ndef mergeKLists_Python2(self, lists):\n\th = []\n\thead = tail = ListNode(0)\n\tfor i in lists:\n\t\tif i:\n\t\t\theapq.heappush(h, (i.val, i))\n\n\twhile h:\n\t\tnode = heapq.heappop(h)[1]\n\t\ttail.next = node\n\t\ttail = tail.next\n\t\tif node.next:\n\t\t\theapq.heappush(h, (node.next.va... | 168 | You are given an array of `k` linked-lists `lists`, each linked-list is sorted in ascending order.
_Merge all the linked-lists into one sorted linked-list and return it._
**Example 1:**
**Input:** lists = \[\[1,4,5\],\[1,3,4\],\[2,6\]\]
**Output:** \[1,1,2,3,4,4,5,6\]
**Explanation:** The linked-lists are:
\[
1->4... | null |
Python solution | heap 96%faster | merge-k-sorted-lists | 0 | 1 | # Code\n```\nclass Solution:\n def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:\n if not lists:\n return\n\n heap = []\n cur = ListNode()\n dummy = cur\n\n for node in lists:\n while node:\n heapq.heappush(heap, node... | 2 | You are given an array of `k` linked-lists `lists`, each linked-list is sorted in ascending order.
_Merge all the linked-lists into one sorted linked-list and return it._
**Example 1:**
**Input:** lists = \[\[1,4,5\],\[1,3,4\],\[2,6\]\]
**Output:** \[1,1,2,3,4,4,5,6\]
**Explanation:** The linked-lists are:
\[
1->4... | null |
Solution | merge-k-sorted-lists | 1 | 1 | ```C++ []\nofstream ans("user.out");\nint main() {\n vector<int> v;\n v.reserve(1e4);\n string s;\n while (getline(cin, s)) {\n s.erase(remove(begin(s), end(s), \'[\'), end(s));\n s.erase(remove(begin(s), end(s), \']\'), end(s));\n for (auto &i: s) if (i == \',\') i = \' \';\n is... | 27 | You are given an array of `k` linked-lists `lists`, each linked-list is sorted in ascending order.
_Merge all the linked-lists into one sorted linked-list and return it._
**Example 1:**
**Input:** lists = \[\[1,4,5\],\[1,3,4\],\[2,6\]\]
**Output:** \[1,1,2,3,4,4,5,6\]
**Explanation:** The linked-lists are:
\[
1->4... | null |
Python || Using While Loop || Easy To Understand 🔥 | merge-k-sorted-lists | 0 | 1 | # Code\n```\nclass Solution:\n def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:\n def mergeTwoLists(left, right) -> Optional[ListNode]:\n if left == None:\n return right\n if right == None:\n return left\n ans = ListNo... | 2 | You are given an array of `k` linked-lists `lists`, each linked-list is sorted in ascending order.
_Merge all the linked-lists into one sorted linked-list and return it._
**Example 1:**
**Input:** lists = \[\[1,4,5\],\[1,3,4\],\[2,6\]\]
**Output:** \[1,1,2,3,4,4,5,6\]
**Explanation:** The linked-lists are:
\[
1->4... | null |
Easy Python Solution [Bruteforce] | merge-k-sorted-lists | 0 | 1 | # Code\n```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\n\nclass Solution:\n def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:\n test = ListNode(0)... | 3 | You are given an array of `k` linked-lists `lists`, each linked-list is sorted in ascending order.
_Merge all the linked-lists into one sorted linked-list and return it._
**Example 1:**
**Input:** lists = \[\[1,4,5\],\[1,3,4\],\[2,6\]\]
**Output:** \[1,1,2,3,4,4,5,6\]
**Explanation:** The linked-lists are:
\[
1->4... | null |
Python || Double Recursion || Clean Code | merge-k-sorted-lists | 0 | 1 | # Intuition + Approach\nThe Solution class has two methods: merge and mergeKLists. The first method, merge, takes two sorted linked lists and merges them into a single sorted linked list in descending order. The second method, mergeKLists, takes a list of k sorted linked lists and recursively divides this list into sma... | 1 | You are given an array of `k` linked-lists `lists`, each linked-list is sorted in ascending order.
_Merge all the linked-lists into one sorted linked-list and return it._
**Example 1:**
**Input:** lists = \[\[1,4,5\],\[1,3,4\],\[2,6\]\]
**Output:** \[1,1,2,3,4,4,5,6\]
**Explanation:** The linked-lists are:
\[
1->4... | null |
✔️ [Python3] PRIORITY QUEUE ೭੧(❛▿❛✿)੭೨, Explained | merge-k-sorted-lists | 0 | 1 | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nThe main challenge in the problem is the fact that the number of lists `k` can be huge and iterating over list heads every time attaching a new node to the result would be time-consuming. Essentially what is importan... | 44 | You are given an array of `k` linked-lists `lists`, each linked-list is sorted in ascending order.
_Merge all the linked-lists into one sorted linked-list and return it._
**Example 1:**
**Input:** lists = \[\[1,4,5\],\[1,3,4\],\[2,6\]\]
**Output:** \[1,1,2,3,4,4,5,6\]
**Explanation:** The linked-lists are:
\[
1->4... | null |
✔️ [Python3] I HATE LINKED LISTS, щ（ﾟДﾟщ）, Not Explained | swap-nodes-in-pairs | 0 | 1 | **UPVOTE if you like (\uD83C\uDF38-_-\'), If you have any question, feel free to ask.**\n\nJust a bunch of placeholders, edge cases, and strange errors about a cycle meanwhile :(\n\nprev cur cur porev next cur prev pasdfaslfjgnzdsf;ljgfsdaz;lkjkfgn\n\nTime: **O(n)** - iterate\nSpace: **O(1)**\n\n```\ndef swapPairs(self... | 174 | Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)
**Example 1:**
**Input:** head = \[1,2,3,4\]
**Output:** \[2,1,4,3\]
**Example 2:**
**Input:** head = \[\]
**Output:** \[\... | null |
Python🔥Java🔥C++🔥Simple Solution🔥Easy to Understand | swap-nodes-in-pairs | 1 | 1 | **!! BIG ANNOUNCEMENT !!**\nI am currently Giving away my premium content well-structured assignments and study materials to clear interviews at top companies related to computer science and data science to my current Subscribers. This is only for first 10,000 Subscribers. **DON\'T FORGET** to Subscribe\n\n# Search \u... | 54 | Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)
**Example 1:**
**Input:** head = \[1,2,3,4\]
**Output:** \[2,1,4,3\]
**Example 2:**
**Input:** head = \[\]
**Output:** \[\... | null |
Python 4-line iterative || Crazy NEXTs | swap-nodes-in-pairs | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nJust for fun. It works.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add y... | 0 | Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)
**Example 1:**
**Input:** head = \[1,2,3,4\]
**Output:** \[2,1,4,3\]
**Example 2:**
**Input:** head = \[\]
**Output:** \[\... | null |
Easy Python Solution || using lists||runtime 97% | swap-nodes-in-pairs | 0 | 1 | \n# Code\n```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution:\n def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:\n lst=[]\n while head:\n lst.appen... | 4 | Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)
**Example 1:**
**Input:** head = \[1,2,3,4\]
**Output:** \[2,1,4,3\]
**Example 2:**
**Input:** head = \[\]
**Output:** \[\... | null |
5 lines of code python O(n) time and space compelxity | swap-nodes-in-pairs | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe plan is to traverse the list by taking two steps at a time.\n\nOn each step, you\'ll swap the node value and till you get to the end of the list. \n\nThis will ignore cases of node that cannot form a pair like [2,3,4], where 4 which i... | 1 | Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)
**Example 1:**
**Input:** head = \[1,2,3,4\]
**Output:** \[2,1,4,3\]
**Example 2:**
**Input:** head = \[\]
**Output:** \[\... | null |
Easy Solution | Beginner Friendly | Easy to Understand | Beats 100% | JAVA | PYTHON 3 |C ++| JS | C# | swap-nodes-in-pairs | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTraverse the list and swap pairs of nodes one by one.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- The node "ans" is to point to the head of the original list. It then uses a "curr" node to traverse the list and... | 24 | Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)
**Example 1:**
**Input:** head = \[1,2,3,4\]
**Output:** \[2,1,4,3\]
**Example 2:**
**Input:** head = \[\]
**Output:** \[\... | null |
✔💯 DAY 411 | BRUTE=>BETTER=>OPTIMAL | 100% | | [PYTHON/JAVA/C++] | EXPLAINED 🆙🆙🆙 | swap-nodes-in-pairs | 1 | 1 | \n\n\n\n# BRUTE\n```JAVA []\npublic ListNode swapPairs(ListNode head) {\n List<ListNode> list = new ArrayList<>();\n ListNode temNode = head;\n while (temNode != null) {\n list.add(temNode);\n temNode = temNode.next;\n }\n for (int i = 0; i < list.size() - 1; i +... | 12 | Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)
**Example 1:**
**Input:** head = \[1,2,3,4\]
**Output:** \[2,1,4,3\]
**Example 2:**
**Input:** head = \[\]
**Output:** \[\... | null |
Python3|| Easy solution || Linkedlist-List. | swap-nodes-in-pairs | 0 | 1 | # Please upvote if you find the solution helful.\n\n# Code\n```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution:\n def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:\n ... | 1 | Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)
**Example 1:**
**Input:** head = \[1,2,3,4\]
**Output:** \[2,1,4,3\]
**Example 2:**
**Input:** head = \[\]
**Output:** \[\... | null |
5 Lines😎|| 0ms ✅|| Easy Approach🔥|| Beasts 100%👌|| Stepped Explanation 😍 | swap-nodes-in-pairs | 1 | 1 | \n# Approach\n<!-- Describe your approach to solving the problem. -->\nHere we are using a very simple and easy ***recursive*** approach to solve this\n\n\n**Steps:**\n\n1. At first we will check, if head value is null we will return null.\n\n `if(head==null) return null;`\n2. Then we check, if next head value is nu... | 7 | Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)
**Example 1:**
**Input:** head = \[1,2,3,4\]
**Output:** \[2,1,4,3\]
**Example 2:**
**Input:** head = \[\]
**Output:** \[\... | null |
✅ C++ | | Python | | Recursive Solution | | Beginners Friendly ✅ | swap-nodes-in-pairs | 0 | 1 | \n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# C++\n```\nclass Solution {\npublic:\n ListNode* swapPairs(ListNode* head) {\n\n if(head == NULL) return NULL;\n ListNod... | 3 | Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)
**Example 1:**
**Input:** head = \[1,2,3,4\]
**Output:** \[2,1,4,3\]
**Example 2:**
**Input:** head = \[\]
**Output:** \[\... | null |
Easy Solution | reverse-nodes-in-k-group | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | Given the `head` of a linked list, reverse the nodes of the list `k` at a time, and return _the modified list_.
`k` is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of `k` then left-out nodes, in the end, should remain as it is.
You may not alt... | null |
Python solution using list | reverse-nodes-in-k-group | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:88%\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$... | 1 | Given the `head` of a linked list, reverse the nodes of the list `k` at a time, and return _the modified list_.
`k` is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of `k` then left-out nodes, in the end, should remain as it is.
You may not alt... | null |
Python Iterative Solution with Constant Space Complexity | reverse-nodes-in-k-group | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe can reverse the one-way nodes using several pointers.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nFirst, we will count the total length of the list.\nSecond, we can now calculate how many groups we should re... | 1 | Given the `head` of a linked list, reverse the nodes of the list `k` at a time, and return _the modified list_.
`k` is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of `k` then left-out nodes, in the end, should remain as it is.
You may not alt... | null |
Python || 97.85% Faster || T.C. O(n) and S.C. O(1) | reverse-nodes-in-k-group | 0 | 1 | ```\nclass Solution:\n def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:\n count=0\n temp=head\n while temp:\n temp=temp.next\n count+=1\n n=count//k #No. of groups to be reversed\n prev=dummy=ListNode()\n dummy.next=head\... | 11 | Given the `head` of a linked list, reverse the nodes of the list `k` at a time, and return _the modified list_.
`k` is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of `k` then left-out nodes, in the end, should remain as it is.
You may not alt... | null |
Clean and Readable Python code | reverse-nodes-in-k-group | 0 | 1 | \n# Code\n```\nclass Solution:\n def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:\n \n dummy = ListNode(0, head)\n currGroup = dummy\n\n while True:\n kth = self.getKth(currGroup, k)\n if not kth:\n break\n ne... | 3 | Given the `head` of a linked list, reverse the nodes of the list `k` at a time, and return _the modified list_.
`k` is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of `k` then left-out nodes, in the end, should remain as it is.
You may not alt... | null |
[Python3] Iterative + recursive solutions || beats 98%🥷🏼 | reverse-nodes-in-k-group | 0 | 1 | # Iterative solution:\n```python3 []\nclass Solution:\n def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:\n def getGroupEnd(cur, k):\n while k > 1 and cur.next:\n cur = cur.next \n k-=1\n return cur if k == 1 else None\n\n ... | 15 | Given the `head` of a linked list, reverse the nodes of the list `k` at a time, and return _the modified list_.
`k` is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of `k` then left-out nodes, in the end, should remain as it is.
You may not alt... | null |
My Solution using Reverse And Link | reverse-nodes-in-k-group | 0 | 1 | \n\n# Code\n```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution:\n def link(self,a,b):\n r=a\n while a.next:\n a=a.next\n a.next=b\n\n def reverseList(self, h... | 4 | Given the `head` of a linked list, reverse the nodes of the list `k` at a time, and return _the modified list_.
`k` is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of `k` then left-out nodes, in the end, should remain as it is.
You may not alt... | null |
python3 easy recursion solution | reverse-nodes-in-k-group | 0 | 1 | ```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\n\nclass Solution:\n def reverseKGroup(self, head: ListNode, k: int) -> ListNode: \n # Check if we need to reverse the group\n curr = he... | 93 | Given the `head` of a linked list, reverse the nodes of the list `k` at a time, and return _the modified list_.
`k` is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of `k` then left-out nodes, in the end, should remain as it is.
You may not alt... | null |
Python solution easy to understand || O(k) space complexity || O(N) time complexity | reverse-nodes-in-k-group | 0 | 1 | ```\nclass Solution:\n def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:\n nodes = []\n cnt = 0\n cpy = head\n while head:\n nodes.append(head)\n head = head.next\n cnt += 1\n if cnt == k:\n cnt = 0\... | 1 | Given the `head` of a linked list, reverse the nodes of the list `k` at a time, and return _the modified list_.
`k` is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of `k` then left-out nodes, in the end, should remain as it is.
You may not alt... | null |
✅ C++ | | Python | | Recursive Solution | | Beginners Friendly ✅ | reverse-nodes-in-k-group | 0 | 1 | # Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: if(recursion stack): O(n) else: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# C++\n```\nclass Solution {\nprivate:\n int length(ListNode* head){\n int size = 0;\n List... | 2 | Given the `head` of a linked list, reverse the nodes of the list `k` at a time, and return _the modified list_.
`k` is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of `k` then left-out nodes, in the end, should remain as it is.
You may not alt... | null |
✅Best Method 🔥|| 100% || C++ || JAVA || PYTHON || Beginner Friendly🔥🔥🔥 | remove-duplicates-from-sorted-array | 1 | 1 | # Intuition:\nThe Intuition is to use two pointers, `i` and `j`, to iterate through the array. The variable `j` is used to keep track of the current index where a unique element should be placed. The initial value of `j` is 1 since the first element in the array is always unique and doesn\'t need to be changed.\n\n# Ex... | 1,168 | Given an integer array `nums` sorted in **non-decreasing order**, remove the duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears only **once**. The **relative order** of the elements should be kept the **same**. Then return _the number of unique elements in_... | In this problem, the key point to focus on is the input array being sorted. As far as duplicate elements are concerned, what is their positioning in the array when the given array is sorted? Look at the image above for the answer. If we know the position of one of the elements, do we also know the positioning of all th... |
[Py]All 4 Methods: Intuitions Walk-through Wrong answer explanations for Beginners [Python]] | remove-duplicates-from-sorted-array | 0 | 1 | \n## \u2705 Method 1: sort in place using `[:]`\n```\n\tdef removeDuplicates(self, nums: List[int]) -> int:\n\t\tnums[:] = sorted(set(nums))\n\t\treturn len(nums)\n```\n\n\n**Time Complexity:** `O(n)`\n**Space Complexity:** `O(1)`\n\n\n### \u274C Common Wrong Answers:\n```\n\tnums = sorted(set(nums))\n\treturn len(nums... | 575 | Given an integer array `nums` sorted in **non-decreasing order**, remove the duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears only **once**. The **relative order** of the elements should be kept the **same**. Then return _the number of unique elements in_... | In this problem, the key point to focus on is the input array being sorted. As far as duplicate elements are concerned, what is their positioning in the array when the given array is sorted? Look at the image above for the answer. If we know the position of one of the elements, do we also know the positioning of all th... |
[VIDEO] Visualization of O(n) Solution | remove-duplicates-from-sorted-array | 0 | 1 | https://www.youtube.com/watch?v=oMr9lehS7Us\n\nOne option is to use the pop() method to pop any duplicates that are detected, but using pop() is actually very inefficient because if we pop an element near the start of the array, we now have to shift all the other elements back 1 space. So each pop() runs in O(n) time,... | 10 | Given an integer array `nums` sorted in **non-decreasing order**, remove the duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears only **once**. The **relative order** of the elements should be kept the **same**. Then return _the number of unique elements in_... | In this problem, the key point to focus on is the input array being sorted. As far as duplicate elements are concerned, what is their positioning in the array when the given array is sorted? Look at the image above for the answer. If we know the position of one of the elements, do we also know the positioning of all th... |
🔥Beats 100% | C++ | Java | Python | Javascript | Ruby | C | O(n) Solution ❣️| Detailed Explanation | remove-duplicates-from-sorted-array | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal is to remove duplicates from a sorted integer array nums while keeping track of the number of unique elements. The function should return the count of unique elements and modify the input array in-place to contain only the unique... | 43 | Given an integer array `nums` sorted in **non-decreasing order**, remove the duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears only **once**. The **relative order** of the elements should be kept the **same**. Then return _the number of unique elements in_... | In this problem, the key point to focus on is the input array being sorted. As far as duplicate elements are concerned, what is their positioning in the array when the given array is sorted? Look at the image above for the answer. If we know the position of one of the elements, do we also know the positioning of all th... |
8 lines two pointer in #python | remove-duplicates-from-sorted-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 169 | Given an integer array `nums` sorted in **non-decreasing order**, remove the duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears only **once**. The **relative order** of the elements should be kept the **same**. Then return _the number of unique elements in_... | In this problem, the key point to focus on is the input array being sorted. As far as duplicate elements are concerned, what is their positioning in the array when the given array is sorted? Look at the image above for the answer. If we know the position of one of the elements, do we also know the positioning of all th... |
Beats 97.98% | Remove Duplicates from Sorted Array...... | remove-duplicates-from-sorted-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- The idea is to use two pointers, one (i) to keep track of the position of the last unique element in the modified array, and another (j) to iterate through the array.\n- When a different element is found, update the modified array by mo... | 4 | Given an integer array `nums` sorted in **non-decreasing order**, remove the duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears only **once**. The **relative order** of the elements should be kept the **same**. Then return _the number of unique elements in_... | In this problem, the key point to focus on is the input array being sorted. As far as duplicate elements are concerned, what is their positioning in the array when the given array is sorted? Look at the image above for the answer. If we know the position of one of the elements, do we also know the positioning of all th... |
Two Pointer | remove-duplicates-from-sorted-array | 0 | 1 | \n# Code\n```\nclass Solution:\n def removeDuplicates(self, nums: List[int]) -> int:\n\n i,j=0,1\n while i<=j and j<len(nums):\n if nums[i]==nums[j]:\n j+=1\n\n else:\n nums[i+1]=nums[j]\n i+=1\n\n return i+1\n``` | 2 | Given an integer array `nums` sorted in **non-decreasing order**, remove the duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears only **once**. The **relative order** of the elements should be kept the **same**. Then return _the number of unique elements in_... | In this problem, the key point to focus on is the input array being sorted. As far as duplicate elements are concerned, what is their positioning in the array when the given array is sorted? Look at the image above for the answer. If we know the position of one of the elements, do we also know the positioning of all th... |
help me modify my code to decrease the runtime | remove-duplicates-from-sorted-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\ni am a beginner and when solving i did use the prebuilt function count() in python to know the number of occurances of the item in the array \n# Approach\n<!-- Describe your approach to solving the problem. -->\ni made use of the count() ... | 0 | Given an integer array `nums` sorted in **non-decreasing order**, remove the duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears only **once**. The **relative order** of the elements should be kept the **same**. Then return _the number of unique elements in_... | In this problem, the key point to focus on is the input array being sorted. As far as duplicate elements are concerned, what is their positioning in the array when the given array is sorted? Look at the image above for the answer. If we know the position of one of the elements, do we also know the positioning of all th... |
Python Easy Solution || 100% | remove-duplicates-from-sorted-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here... | 1 | Given an integer array `nums` sorted in **non-decreasing order**, remove the duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears only **once**. The **relative order** of the elements should be kept the **same**. Then return _the number of unique elements in_... | In this problem, the key point to focus on is the input array being sorted. As far as duplicate elements are concerned, what is their positioning in the array when the given array is sorted? Look at the image above for the answer. If we know the position of one of the elements, do we also know the positioning of all th... |
EASY PYTHON SOLUTION | remove-duplicates-from-sorted-array | 0 | 1 | ```\nclass Solution:\n def removeDuplicates(self, nums: List[int]) -> int:\n if len(nums)==1:\n return 1\n x=nums[0]\n i=0\n while i<len(nums)-1:\n if x==nums[1+i]:\n del nums[1+i]\n else:\n x=nums[1+i]\n i+=1\n... | 8 | Given an integer array `nums` sorted in **non-decreasing order**, remove the duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) such that each unique element appears only **once**. The **relative order** of the elements should be kept the **same**. Then return _the number of unique elements in_... | In this problem, the key point to focus on is the input array being sorted. As far as duplicate elements are concerned, what is their positioning in the array when the given array is sorted? Look at the image above for the answer. If we know the position of one of the elements, do we also know the positioning of all th... |
✅Best 100% || C++ || JAVA || PYTHON || Beginner Friendly🔥🔥🔥 | remove-element | 1 | 1 | # Intuition\r\n<!-- Describe your first thoughts on how to solve this problem. -->\r\nThe intuition behind this solution is to iterate through the array and keep track of two pointers: `index` and `i`. The `index` pointer represents the position where the next non-target element should be placed, while the `i` pointer ... | 1,020 | Given an integer array `nums` and an integer `val`, remove all occurrences of `val` in `nums` [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm). The order of the elements may be changed. Then return _the number of elements in_ `nums` _which are not equal to_ `val`.
Consider the number of elements in `nu... | The problem statement clearly asks us to modify the array in-place and it also says that the element beyond the new length of the array can be anything. Given an element, we need to remove all the occurrences of it from the array. We don't technically need to remove that element per-say, right? We can move all the occu... |
2nd | remove-element | 0 | 1 | 2\n# Code\n```\nclass Solution:\n def removeElement(self, nums: List[int], val: int) -> int:\n k = len(nums)\n i = 0\n while i < k:\n if nums[i] == val:\n nums[i] = nums[k-1]\n nums[k-1] = \'_\'\n k -= 1\n i -= 1\n ... | 1 | Given an integer array `nums` and an integer `val`, remove all occurrences of `val` in `nums` [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm). The order of the elements may be changed. Then return _the number of elements in_ `nums` _which are not equal to_ `val`.
Consider the number of elements in `nu... | The problem statement clearly asks us to modify the array in-place and it also says that the element beyond the new length of the array can be anything. Given an element, we need to remove all the occurrences of it from the array. We don't technically need to remove that element per-say, right? We can move all the occu... |
[VIDEO] Clear Visualization of O(n) Solution | remove-element | 0 | 1 | https://youtu.be/pGKDzt0gk-A\n\nOne way to do this is to check every element of `nums`, and if an element is equal to `val`, then we use the pop method to pop it off. This works, but the problem is that the pop method is very inefficient because if we pop an element near the start of the array, all the other elements h... | 12 | Given an integer array `nums` and an integer `val`, remove all occurrences of `val` in `nums` [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm). The order of the elements may be changed. Then return _the number of elements in_ `nums` _which are not equal to_ `val`.
Consider the number of elements in `nu... | The problem statement clearly asks us to modify the array in-place and it also says that the element beyond the new length of the array can be anything. Given an element, we need to remove all the occurrences of it from the array. We don't technically need to remove that element per-say, right? We can move all the occu... |
2 Lines of Code Python3 | remove-element | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 117 | Given an integer array `nums` and an integer `val`, remove all occurrences of `val` in `nums` [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm). The order of the elements may be changed. Then return _the number of elements in_ `nums` _which are not equal to_ `val`.
Consider the number of elements in `nu... | The problem statement clearly asks us to modify the array in-place and it also says that the element beyond the new length of the array can be anything. Given an element, we need to remove all the occurrences of it from the array. We don't technically need to remove that element per-say, right? We can move all the occu... |
✔️💯% || Easy explanation || Beginner Friendly🔥🔥 | remove-element | 1 | 1 | # Intuition\nThe problem asks to remove all occurrences of a specific value (val) in the given array (nums) in-place. Additionally, it requires returning the count of elements in the modified array that are not equal to the given value. The order of the elements may change after the removal.\n\n# Approach\nWe can solve... | 3 | Given an integer array `nums` and an integer `val`, remove all occurrences of `val` in `nums` [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm). The order of the elements may be changed. Then return _the number of elements in_ `nums` _which are not equal to_ `val`.
Consider the number of elements in `nu... | The problem statement clearly asks us to modify the array in-place and it also says that the element beyond the new length of the array can be anything. Given an element, we need to remove all the occurrences of it from the array. We don't technically need to remove that element per-say, right? We can move all the occu... |
simple easy solution ||beats 93% 🤖💻🧑💻 | remove-element | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | Given an integer array `nums` and an integer `val`, remove all occurrences of `val` in `nums` [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm). The order of the elements may be changed. Then return _the number of elements in_ `nums` _which are not equal to_ `val`.
Consider the number of elements in `nu... | The problem statement clearly asks us to modify the array in-place and it also says that the element beyond the new length of the array can be anything. Given an element, we need to remove all the occurrences of it from the array. We don't technically need to remove that element per-say, right? We can move all the occu... |
Python two pointer| Beats 99% | remove-element | 0 | 1 | ```\nGiven : val = 2\n1 2 2 2 5 2 5 \ni j\nFirst Start from the far end \n1 2 2 2 5 2 5 \n i j\nwhen nums[i] == val , nums[i] = nums[j]\n1 2 2 2 5 2 5 \n i j \nif(num[j]) is also val then move it left till its not\n1 5 5 2 2 2 5 \n i j \nwhen i==j.. stop\n1 5 5 2 2 2 5 \n i,j\nyou... | 8 | Given an integer array `nums` and an integer `val`, remove all occurrences of `val` in `nums` [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm). The order of the elements may be changed. Then return _the number of elements in_ `nums` _which are not equal to_ `val`.
Consider the number of elements in `nu... | The problem statement clearly asks us to modify the array in-place and it also says that the element beyond the new length of the array can be anything. Given an element, we need to remove all the occurrences of it from the array. We don't technically need to remove that element per-say, right? We can move all the occu... |
Two approach for easy understanding in Python | With and Without python functions | remove-element | 0 | 1 | # Approach1\n## Time complexity: O(nlogn)\n## Space Complexity = O(1)\n\n# Approach2\n## Time complexity: O(n)\n## Space Complexity = O(1)\n\n# Code\n``` Approach1 []\n\n# Iterate linearly over the nums array.\n# Find the nums with val and replace it with something else.\n# Sort the elements and return the count \n\ncl... | 2 | Given an integer array `nums` and an integer `val`, remove all occurrences of `val` in `nums` [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm). The order of the elements may be changed. Then return _the number of elements in_ `nums` _which are not equal to_ `val`.
Consider the number of elements in `nu... | The problem statement clearly asks us to modify the array in-place and it also says that the element beyond the new length of the array can be anything. Given an element, we need to remove all the occurrences of it from the array. We don't technically need to remove that element per-say, right? We can move all the occu... |
a brilliant solution from an outstanding and gorgeous scientist | remove-element | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | Given an integer array `nums` and an integer `val`, remove all occurrences of `val` in `nums` [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm). The order of the elements may be changed. Then return _the number of elements in_ `nums` _which are not equal to_ `val`.
Consider the number of elements in `nu... | The problem statement clearly asks us to modify the array in-place and it also says that the element beyond the new length of the array can be anything. Given an element, we need to remove all the occurrences of it from the array. We don't technically need to remove that element per-say, right? We can move all the occu... |
Removing by sorting (welcome to ask questions :)) | remove-element | 0 | 1 | # Intuition\nSort elemts that are equal to val and remove them\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nSort elements that are equal to val and leave that are not equal\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add ... | 3 | Given an integer array `nums` and an integer `val`, remove all occurrences of `val` in `nums` [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm). The order of the elements may be changed. Then return _the number of elements in_ `nums` _which are not equal to_ `val`.
Consider the number of elements in `nu... | The problem statement clearly asks us to modify the array in-place and it also says that the element beyond the new length of the array can be anything. Given an element, we need to remove all the occurrences of it from the array. We don't technically need to remove that element per-say, right? We can move all the occu... |
[ PYTHON ] Simple Two Pointer Solution Beats 95% || | remove-element | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 5 | Given an integer array `nums` and an integer `val`, remove all occurrences of `val` in `nums` [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm). The order of the elements may be changed. Then return _the number of elements in_ `nums` _which are not equal to_ `val`.
Consider the number of elements in `nu... | The problem statement clearly asks us to modify the array in-place and it also says that the element beyond the new length of the array can be anything. Given an element, we need to remove all the occurrences of it from the array. We don't technically need to remove that element per-say, right? We can move all the occu... |
Beat 100% Python One Line, Most Pythonic and List Comprehension No While Loops | remove-element | 0 | 1 | # Intuition\nSeems like any other pythonic list comprehension problem. Only hurdle is that it has to be in place\n# Approach\nBy using nums[:] you can make it in place\n# Complexity\n- Time complexity:\n\n- Space complexity:\n\n# Code\n```\nclass Solution:\n def removeElement(self, nums: List[int], val: int) -> int:... | 8 | Given an integer array `nums` and an integer `val`, remove all occurrences of `val` in `nums` [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm). The order of the elements may be changed. Then return _the number of elements in_ `nums` _which are not equal to_ `val`.
Consider the number of elements in `nu... | The problem statement clearly asks us to modify the array in-place and it also says that the element beyond the new length of the array can be anything. Given an element, we need to remove all the occurrences of it from the array. We don't technically need to remove that element per-say, right? We can move all the occu... |
Two-Pointer In-Place Removal Solution. | remove-element | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this approach is to iterate through the `nums` array and keep track of the number of elements not equal to `val`. We will move the elements not equal to `val` to the front of the array and update the count accordingly... | 1 | Given an integer array `nums` and an integer `val`, remove all occurrences of `val` in `nums` [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm). The order of the elements may be changed. Then return _the number of elements in_ `nums` _which are not equal to_ `val`.
Consider the number of elements in `nu... | The problem statement clearly asks us to modify the array in-place and it also says that the element beyond the new length of the array can be anything. Given an element, we need to remove all the occurrences of it from the array. We don't technically need to remove that element per-say, right? We can move all the occu... |
Python - Remove Element - just 6 lines of code | remove-element | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | Given an integer array `nums` and an integer `val`, remove all occurrences of `val` in `nums` [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm). The order of the elements may be changed. Then return _the number of elements in_ `nums` _which are not equal to_ `val`.
Consider the number of elements in `nu... | The problem statement clearly asks us to modify the array in-place and it also says that the element beyond the new length of the array can be anything. Given an element, we need to remove all the occurrences of it from the array. We don't technically need to remove that element per-say, right? We can move all the occu... |
⭐C#, Java, Python3,JavaScript Solution (Faster than 90~100%) | remove-element | 1 | 1 | \nSee the code : **\u2B50[https://zyrastory.com/en/coding-en/leetcode-en/leetcode-27-remove-element-from-sorted-array-solution-and-explanation-en/](https://zyrastory.com/en/coding-en/leetcode-en/leetcode-27-remove-element-from-sorted-array-solution-and-explanation-en/)\u2B50**\n\n**Submissioon detail**\n. The order of the elements may be changed. Then return _the number of elements in_ `nums` _which are not equal to_ `val`.
Consider the number of elements in `nu... | The problem statement clearly asks us to modify the array in-place and it also says that the element beyond the new length of the array can be anything. Given an element, we need to remove all the occurrences of it from the array. We don't technically need to remove that element per-say, right? We can move all the occu... |
Simple and Easy Solution🔥🔥|| Explained for finding first occurrence of string ||Beats 100%🔥🔥 | find-the-index-of-the-first-occurrence-in-a-string | 0 | 1 | # Intuition\nThe problem calls for finding the first occurrence of a substring (needle) within another string (haystack). If the needle is not present in the haystack, the function should return -1.\n# Approach\n1. Initialization: Get lengths of needle (m) and haystack (n).\n2. Equal Strings Check: If haystack equals n... | 2 | Given two strings `needle` and `haystack`, return the index of the first occurrence of `needle` in `haystack`, or `-1` if `needle` is not part of `haystack`.
**Example 1:**
**Input:** haystack = "sadbutsad ", needle = "sad "
**Output:** 0
**Explanation:** "sad " occurs at index 0 and 6.
The first occurrence is at ... | null |
[VIDEO] Visualization of O(n) KMP Algorithm | find-the-index-of-the-first-occurrence-in-a-string | 0 | 1 | https://youtu.be/0iDiUuHZE_g\n\nThe KMP algorithm improves the brute force algorithm from O(m*n) to O(m+n) and consists of two parts. The first part is the preprocessing step, where we find the length of "the longest proper prefix that is also a suffix" for every prefix in `needle`.\n\nThe second part is the main algo... | 15 | Given two strings `needle` and `haystack`, return the index of the first occurrence of `needle` in `haystack`, or `-1` if `needle` is not part of `haystack`.
**Example 1:**
**Input:** haystack = "sadbutsad ", needle = "sad "
**Output:** 0
**Explanation:** "sad " occurs at index 0 and 6.
The first occurrence is at ... | null |
Easiest, Big IQ, fast solution | Beats 97% of solutions | find-the-index-of-the-first-occurrence-in-a-string | 0 | 1 | \n# Code\n```\nclass Solution(object):\n def strStr(self, haystack, needle):\n # makes sure we don\'t iterate through a substring that is shorter than needle\n for i in range(len(haystack) - len(needle) + 1):\n # check if any substring of haystack with the same length as needle is equal to n... | 28 | Given two strings `needle` and `haystack`, return the index of the first occurrence of `needle` in `haystack`, or `-1` if `needle` is not part of `haystack`.
**Example 1:**
**Input:** haystack = "sadbutsad ", needle = "sad "
**Output:** 0
**Explanation:** "sad " occurs at index 0 and 6.
The first occurrence is at ... | null |
one line code using python ( TC: O(N) | SC: O(1)) | find-the-index-of-the-first-occurrence-in-a-string | 0 | 1 | # Intuition\nThe intuition behind this approach is to use the built-in find() function in Python to find the index of the first occurrence of the needle string within the haystack string. If the needle is not found, it returns -1.\n# Approach\nThe approach is simple and straightforward. We can directly use the find() f... | 47 | Given two strings `needle` and `haystack`, return the index of the first occurrence of `needle` in `haystack`, or `-1` if `needle` is not part of `haystack`.
**Example 1:**
**Input:** haystack = "sadbutsad ", needle = "sad "
**Output:** 0
**Explanation:** "sad " occurs at index 0 and 6.
The first occurrence is at ... | null |
Find the Index of the First Occurrence in a String - 3 | find-the-index-of-the-first-occurrence-in-a-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | Given two strings `needle` and `haystack`, return the index of the first occurrence of `needle` in `haystack`, or `-1` if `needle` is not part of `haystack`.
**Example 1:**
**Input:** haystack = "sadbutsad ", needle = "sad "
**Output:** 0
**Explanation:** "sad " occurs at index 0 and 6.
The first occurrence is at ... | null |
EASIEST PYTHON SOLUTION BEATS 99.6% MEMORY | find-the-index-of-the-first-occurrence-in-a-string | 0 | 1 | \n# Code\n```\nclass Solution:\n def strStr(self, haystack: str, needle: str) -> int:\n try: return haystack.find(needle) \n except: return -1\n``` | 1 | Given two strings `needle` and `haystack`, return the index of the first occurrence of `needle` in `haystack`, or `-1` if `needle` is not part of `haystack`.
**Example 1:**
**Input:** haystack = "sadbutsad ", needle = "sad "
**Output:** 0
**Explanation:** "sad " occurs at index 0 and 6.
The first occurrence is at ... | null |
Python String Search | ".find" & ".index" | find-the-index-of-the-first-occurrence-in-a-string | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe need to find the index of the first occurrence of the target string in the given input string if it is a substring.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nString search logic:\n1. Check if the target st... | 3 | Given two strings `needle` and `haystack`, return the index of the first occurrence of `needle` in `haystack`, or `-1` if `needle` is not part of `haystack`.
**Example 1:**
**Input:** haystack = "sadbutsad ", needle = "sad "
**Output:** 0
**Explanation:** "sad " occurs at index 0 and 6.
The first occurrence is at ... | null |
Easy Solution in Python3 | divide-two-integers | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | Given two integers `dividend` and `divisor`, divide two integers **without** using multiplication, division, and mod operator.
The integer division should truncate toward zero, which means losing its fractional part. For example, `8.345` would be truncated to `8`, and `-2.7335` would be truncated to `-2`.
Return _the... | null |
Python3 - easiest solution | divide-two-integers | 0 | 1 | 1) Define the result\'s sign and operate with positive dividend and divisor.\n2) Calculate the result using the length of range.\n3) Apply the sign.\n4) Apply the 32-bit integer limitations.\n\nNo multiplication, division, or mod used.\n\n\n```\nclass Solution:\n def divide(self, dividend: int, divisor: int) -> int:... | 41 | Given two integers `dividend` and `divisor`, divide two integers **without** using multiplication, division, and mod operator.
The integer division should truncate toward zero, which means losing its fractional part. For example, `8.345` would be truncated to `8`, and `-2.7335` would be truncated to `-2`.
Return _the... | null |
Bits of Python | divide-two-integers | 0 | 1 | \n# Code\n```\nclass Solution:\n def divide(self, dividend: int, divisor: int) -> int:\n sign = -1 if (dividend >= 0 and divisor < 0) or (dividend < 0 and divisor >= 0) else 1\n dividend = abs(dividend)\n divisor = abs(divisor)\n result = len(range(0, dividend-divisor+1, divisor))\n ... | 4 | Given two integers `dividend` and `divisor`, divide two integers **without** using multiplication, division, and mod operator.
The integer division should truncate toward zero, which means losing its fractional part. For example, `8.345` would be truncated to `8`, and `-2.7335` would be truncated to `-2`.
Return _the... | null |
Efficient Bit Manipulation Algorithm for Integer Division without Multiplication or Mod Operator. | divide-two-integers | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires us to divide two integers, `dividend` and `divisor`, and return the quotient without using multiplication, division, and mod operator.\n\nWe can solve this problem using bit manipulation and binary search. We can find... | 17 | Given two integers `dividend` and `divisor`, divide two integers **without** using multiplication, division, and mod operator.
The integer division should truncate toward zero, which means losing its fractional part. For example, `8.345` would be truncated to `8`, and `-2.7335` would be truncated to `-2`.
Return _the... | null |
My solution for this problem | divide-two-integers | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n# Complexity\n- Time complexity:O[log(y)]\n<!-- Add your time complexity here , e.g. $$O(n)$$ -->\n\n- Space complexity:O[1]\n<!-- Add your space complexity here, e.g.... | 0 | Given two integers `dividend` and `divisor`, divide two integers **without** using multiplication, division, and mod operator.
The integer division should truncate toward zero, which means losing its fractional part. For example, `8.345` would be truncated to `8`, and `-2.7335` would be truncated to `-2`.
Return _the... | null |
Python fast code with detailed explanation | divide-two-integers | 0 | 1 | I took the code of [@tusizi](https://leetcode.com/problems/divide-two-integers/discuss/13403/Clear-python-code) and made it easier to understand by changing variable names, and adding a detailed explanation for each line. I\'m also not experienced myself and whatever I right here comes from discussion forums and googl... | 42 | Given two integers `dividend` and `divisor`, divide two integers **without** using multiplication, division, and mod operator.
The integer division should truncate toward zero, which means losing its fractional part. For example, `8.345` would be truncated to `8`, and `-2.7335` would be truncated to `-2`.
Return _the... | null |
Python Subtraction O(logN) | divide-two-integers | 0 | 1 | You can refer to this [video solution](https://www.youtube.com/watch?v=xefkgtd44hg&ab_channel=CheatCodeNinja) \n\nHere we are doubling the divisor everytime and hence compressing the total number of subtraction operations. \nIf the doubled divisor is greater than the dividend, we reset our divisor to its initial value ... | 12 | Given two integers `dividend` and `divisor`, divide two integers **without** using multiplication, division, and mod operator.
The integer division should truncate toward zero, which means losing its fractional part. For example, `8.345` would be truncated to `8`, and `-2.7335` would be truncated to `-2`.
Return _the... | null |
Python solution using hash map | substring-with-concatenation-of-all-words | 0 | 1 | \n# Code\n```\nclass Solution:\n\n def calc(self ,i):\n\n cnt = 0\n ind = i\n while (ind < i+self.pl):\n news = self.s[ind : ind + self.n]\n if news in self.dic :\n self.dic[news] -= 1 \n if self.dic[news] == 0 :\n cnt += 1 \... | 1 | You are given a string `s` and an array of strings `words`. All the strings of `words` are of **the same length**.
A **concatenated substring** in `s` is a substring that contains all the strings of any permutation of `words` concatenated.
* For example, if `words = [ "ab ", "cd ", "ef "]`, then `"abcdef "`, `"abef... | null |
Simple sliding window solution (beats 99% for time) | substring-with-concatenation-of-all-words | 0 | 1 | \n\n# Approach\nTo solve this problem, we can use a sliding window approach combined with a hashmap to keep track of the word counts. The steps are as follows:\n\n1. Calculate Word Lengths and Total Length: Since all words in words are of the same length, calculate the length of each word and the total length of the co... | 2 | You are given a string `s` and an array of strings `words`. All the strings of `words` are of **the same length**.
A **concatenated substring** in `s` is a substring that contains all the strings of any permutation of `words` concatenated.
* For example, if `words = [ "ab ", "cd ", "ef "]`, then `"abcdef "`, `"abef... | null |
Solution | substring-with-concatenation-of-all-words | 1 | 1 | ```C++ []\nclass Solution {\npublic:\n\tstruct matcher {\n\t\tstruct info { int mtindex, count; };\n\t\tunordered_map<string_view, info>dict;\n\t\tint different_word_count;\n\n\t\tvector<int>slot;\n\t\tint maching_slot_count;\n\n\t\tmatcher(const vector<string>& words) {\n\t\t\tint mtind = 0;\n\t\t\tfor (auto& word : w... | 28 | You are given a string `s` and an array of strings `words`. All the strings of `words` are of **the same length**.
A **concatenated substring** in `s` is a substring that contains all the strings of any permutation of `words` concatenated.
* For example, if `words = [ "ab ", "cd ", "ef "]`, then `"abcdef "`, `"abef... | null |
✅ Explained - Simple and Clear Python3 Code✅ | substring-with-concatenation-of-all-words | 0 | 1 | \n# Approach\n\nThe solution iterates through possible starting points in the given string and checks if the substring starting from each point forms a concatenated substring. It maintains a dictionary to track the occurrences of words encountered and compares it with the expected occurrences. By updating the counts wh... | 6 | You are given a string `s` and an array of strings `words`. All the strings of `words` are of **the same length**.
A **concatenated substring** in `s` is a substring that contains all the strings of any permutation of `words` concatenated.
* For example, if `words = [ "ab ", "cd ", "ef "]`, then `"abcdef "`, `"abef... | null |
Substring with Concatenation of All Words with step by step explanation, | substring-with-concatenation-of-all-words | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nOne approach to solve this problem is using a sliding window technique.\n\nStep 1: Calculate the length of each word and the total length of words.\n\nStep 2: Create a dictionary to store the frequency of each word in words.... | 1 | You are given a string `s` and an array of strings `words`. All the strings of `words` are of **the same length**.
A **concatenated substring** in `s` is a substring that contains all the strings of any permutation of `words` concatenated.
* For example, if `words = [ "ab ", "cd ", "ef "]`, then `"abcdef "`, `"abef... | null |
Python Elegant & Short | Sliding window | 98.89% faster | substring-with-concatenation-of-all-words | 0 | 1 | \n\n```\nfrom collections import Counter, defaultdict\n\n\nclass Solution:\n\t"""\n\tTime: O(n*k), n = length of s, k = length of each word\n\tMemory: O(m*k), m = length of words, k = length of each word\n\t""... | 16 | You are given a string `s` and an array of strings `words`. All the strings of `words` are of **the same length**.
A **concatenated substring** in `s` is a substring that contains all the strings of any permutation of `words` concatenated.
* For example, if `words = [ "ab ", "cd ", "ef "]`, then `"abcdef "`, `"abef... | null |
SUBSTRING WITH CONCATENATION of all words | substring-with-concatenation-of-all-words | 0 | 1 | [https://leetcode.com/problems/substring-with-concatenation-of-all-words/description/]()\n[https://www.linkedin.com/in/shivansh-srivastava-cs/]()\n# DO UPVOTE IF YOU FIND IT SUITABLE AND HELPFUL AND FOR ANY DOUBTS COMMENT.\n# DO LIKE AS EACH VOTE COUNTS\n\n\n# Intuition\n<!-- Describe your first thoughts on how to solv... | 6 | You are given a string `s` and an array of strings `words`. All the strings of `words` are of **the same length**.
A **concatenated substring** in `s` is a substring that contains all the strings of any permutation of `words` concatenated.
* For example, if `words = [ "ab ", "cd ", "ef "]`, then `"abcdef "`, `"abef... | null |
Simple python solution | substring-with-concatenation-of-all-words | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | You are given a string `s` and an array of strings `words`. All the strings of `words` are of **the same length**.
A **concatenated substring** in `s` is a substring that contains all the strings of any permutation of `words` concatenated.
* For example, if `words = [ "ab ", "cd ", "ef "]`, then `"abcdef "`, `"abef... | null |
Python full two step easy explanation with examples | next-permutation | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nLike Djikstra algorithm, finding the next permutation is something that its better to just know the steps. Knowing the steps is key which I illustrated in the comments. Now I think the challenge is if the interviewer told you the solution... | 3 | A **permutation** of an array of integers is an arrangement of its members into a sequence or linear order.
* For example, for `arr = [1,2,3]`, the following are all the permutations of `arr`: `[1,2,3], [1,3,2], [2, 1, 3], [2, 3, 1], [3,1,2], [3,2,1]`.
The **next permutation** of an array of integers is the next le... | null |
Python3 | O(n) | next-permutation | 0 | 1 | ```\nclass Solution:\n def nextPermutation(self, nums: List[int]) -> None:\n """\n Do not return anything, modify nums in-place instead.\n """\n for curr in reversed(range(0, len(nums) - 1)):\n if nums[curr] >= nums[curr + 1]:\n continue;\n \n ... | 2 | A **permutation** of an array of integers is an arrangement of its members into a sequence or linear order.
* For example, for `arr = [1,2,3]`, the following are all the permutations of `arr`: `[1,2,3], [1,3,2], [2, 1, 3], [2, 3, 1], [3,1,2], [3,2,1]`.
The **next permutation** of an array of integers is the next le... | null |
Python 3 Solution beats 96.6% | next-permutation | 0 | 1 | \n# Approach\n<!-- Describe your approach to solving the problem. -->\nAlgorithm / Intuition\nThe steps are the following:\n\n1.) Find the break-point, i: Break-point means the first index i from the back of the given array where arr[i] becomes smaller than arr[i+1].\nFor example, if the given array is {2,1,5,4,3,0,0},... | 2 | A **permutation** of an array of integers is an arrangement of its members into a sequence or linear order.
* For example, for `arr = [1,2,3]`, the following are all the permutations of `arr`: `[1,2,3], [1,3,2], [2, 1, 3], [2, 3, 1], [3,1,2], [3,2,1]`.
The **next permutation** of an array of integers is the next le... | null |
Great Python Solution! Using 'Sort' and 'UpperBound' | O(NlogN) Time complexity | O(1) Space | next-permutation | 0 | 1 | # Intuition & Approach\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe have to find the first index where the ascending starts Ex. in 2,1,3,2 element at index 2 is greater than element at index 1 hence index 1 is the target index. Now we have to replace that index with the upper bound of the el... | 3 | A **permutation** of an array of integers is an arrangement of its members into a sequence or linear order.
* For example, for `arr = [1,2,3]`, the following are all the permutations of `arr`: `[1,2,3], [1,3,2], [2, 1, 3], [2, 3, 1], [3,1,2], [3,2,1]`.
The **next permutation** of an array of integers is the next le... | null |
Python Solution with easy understanding and comments. (Bonus: Related links) | next-permutation | 0 | 1 | ```\nclass Solution:\n def nextPermutation(self, nums: List[int]) -> None:\n """\n Do not return anything, modify nums in-place instead.\n """\n # To find next permutations, we\'ll start from the end\n i = j = len(nums)-1\n # First we\'ll find the first non-increasing elemen... | 88 | A **permutation** of an array of integers is an arrangement of its members into a sequence or linear order.
* For example, for `arr = [1,2,3]`, the following are all the permutations of `arr`: `[1,2,3], [1,3,2], [2, 1, 3], [2, 3, 1], [3,1,2], [3,2,1]`.
The **next permutation** of an array of integers is the next le... | null |
Easy To Understand | longest-valid-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 4 | Given a string containing just the characters `'('` and `')'`, return _the length of the longest valid (well-formed) parentheses_ _substring_.
**Example 1:**
**Input:** s = "(() "
**Output:** 2
**Explanation:** The longest valid parentheses substring is "() ".
**Example 2:**
**Input:** s = ")()()) "
**Output:** ... | null |
Python Easy 2 approaches - Space O(n) and O(1) | longest-valid-parentheses | 0 | 1 | The time complexity for both the approaches is `O(len(s)`.\n\n 1. ##### **Space - O(len(s))**\n\nThis approach solves the problem in similar way as https://leetcode.com/problems/valid-parentheses/ using `Stack`. The stack is used to track indices of `(`. So whenever we hit a `)`, we pop the pair from stack and update t... | 61 | Given a string containing just the characters `'('` and `')'`, return _the length of the longest valid (well-formed) parentheses_ _substring_.
**Example 1:**
**Input:** s = "(() "
**Output:** 2
**Explanation:** The longest valid parentheses substring is "() ".
**Example 2:**
**Input:** s = ")()()) "
**Output:** ... | null |
Python 56ms Solution | longest-valid-parentheses | 0 | 1 | \n<!-- Describe your first thoughts on how to solve this problem. -->\n1. Firstly determine which of the consecutive parentheses are valid and create a duplicate list of the given string where each bracket is a different element i.e.\nif $$givenstring="( ( )"$$\ncreate list =list(given_string) which will result in$$ [ ... | 1 | Given a string containing just the characters `'('` and `')'`, return _the length of the longest valid (well-formed) parentheses_ _substring_.
**Example 1:**
**Input:** s = "(() "
**Output:** 2
**Explanation:** The longest valid parentheses substring is "() ".
**Example 2:**
**Input:** s = ")()()) "
**Output:** ... | null |
✅Python3 Beats 99.80% Solution with Detailed Explanation✅ | longest-valid-parentheses | 0 | 1 | # Approach\n1. Initiate stack [-1] (Must be -1 !!)\n2. If you cannot understand why it must be -1, consider the case "(()())"\n3. The -1 is to get the correct current position in case there is nothing to pop in the stack\n4. For every element, if it is "(", we append the current index\n5. *Append index to keep track of... | 1 | Given a string containing just the characters `'('` and `')'`, return _the length of the longest valid (well-formed) parentheses_ _substring_.
**Example 1:**
**Input:** s = "(() "
**Output:** 2
**Explanation:** The longest valid parentheses substring is "() ".
**Example 2:**
**Input:** s = ")()()) "
**Output:** ... | null |
Dynamic Programming Approach to Longest Valid Parentheses Substring | longest-valid-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo solve this problem, we can use dynamic programming. Let dp[i] be the length of the longest valid parentheses substring ending at index i. We can observe the following:\n\n1. If s[i] = \'(\', there can be no valid substring ending at in... | 4 | Given a string containing just the characters `'('` and `')'`, return _the length of the longest valid (well-formed) parentheses_ _substring_.
**Example 1:**
**Input:** s = "(() "
**Output:** 2
**Explanation:** The longest valid parentheses substring is "() ".
**Example 2:**
**Input:** s = ")()()) "
**Output:** ... | null |
Longest Valid Parenthesesis (Valid Parenthesesis) | longest-valid-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIt is similar to the valid parenthesesis but there we check the validity condition but in this we need to find the length of the longest valid parenthesis so inorder to find the length we needed to store the index of the brackets and trav... | 2 | Given a string containing just the characters `'('` and `')'`, return _the length of the longest valid (well-formed) parentheses_ _substring_.
**Example 1:**
**Input:** s = "(() "
**Output:** 2
**Explanation:** The longest valid parentheses substring is "() ".
**Example 2:**
**Input:** s = ")()()) "
**Output:** ... | null |
Easiest Python3 Solution [ Beats 97.40% ] 🔥🔥 | longest-valid-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- The intuition behind solving this problem is to use a stack data structure to keep track of the indices of opening parentheses \'(\' in the input string. By doing this, we can identify valid pairs of parentheses and calculate their leng... | 2 | Given a string containing just the characters `'('` and `')'`, return _the length of the longest valid (well-formed) parentheses_ _substring_.
**Example 1:**
**Input:** s = "(() "
**Output:** 2
**Explanation:** The longest valid parentheses substring is "() ".
**Example 2:**
**Input:** s = ")()()) "
**Output:** ... | null |
Stack Approach .. Using Python | longest-valid-parentheses | 0 | 1 | # Code\n```\nclass Solution:\n def longestValidParentheses(self, s: str) -> int:\n n = len(s)\n stack = []\n\n stack += [["first_node" , -1]]\n \n for i in range(n):\n if s[i] == "(": \n stack += [["(" , i]]\n \n else: # s[i] ==... | 2 | Given a string containing just the characters `'('` and `')'`, return _the length of the longest valid (well-formed) parentheses_ _substring_.
**Example 1:**
**Input:** s = "(() "
**Output:** 2
**Explanation:** The longest valid parentheses substring is "() ".
**Example 2:**
**Input:** s = ")()()) "
**Output:** ... | null |
Python3 Simple Solution | longest-valid-parentheses | 0 | 1 | ```\nclass Solution:\n def longestValidParentheses(self, s: str) -> int:\n \n maxi = 0\n stack = [-1]\n \n for i in range(len(s)) :\n if s[i] == "(" : stack.append(i)\n else :\n stack.pop()\n if len(stack) == 0 : stack.append(i)\n... | 4 | Given a string containing just the characters `'('` and `')'`, return _the length of the longest valid (well-formed) parentheses_ _substring_.
**Example 1:**
**Input:** s = "(() "
**Output:** 2
**Explanation:** The longest valid parentheses substring is "() ".
**Example 2:**
**Input:** s = ")()()) "
**Output:** ... | null |
[Python 93%] Easy explained short solution with stack | longest-valid-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWhile iterating through the string, we can use a stack to keep track of the indices of the opening parentheses that have not yet been matched with a closing parenthesis. When we encounter a closing parenthesis, we pop the last element off... | 1 | Given a string containing just the characters `'('` and `')'`, return _the length of the longest valid (well-formed) parentheses_ _substring_.
**Example 1:**
**Input:** s = "(() "
**Output:** 2
**Explanation:** The longest valid parentheses substring is "() ".
**Example 2:**
**Input:** s = ")()()) "
**Output:** ... | null |
Easy Python Code | longest-valid-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(n)\n<!-- Add your space complexity here, e.g. $$... | 0 | Given a string containing just the characters `'('` and `')'`, return _the length of the longest valid (well-formed) parentheses_ _substring_.
**Example 1:**
**Input:** s = "(() "
**Output:** 2
**Explanation:** The longest valid parentheses substring is "() ".
**Example 2:**
**Input:** s = ")()()) "
**Output:** ... | null |
Two Approaches Using Stack and without space | longest-valid-parentheses | 0 | 1 | # Using Stack:\n```\nclass Solution:\n def longestValidParentheses(self, s: str) -> int:\n stack=[]\n stack.append(-1)\n ans=0\n for i in range(len(s)):\n if s[i]==\'(\':\n stack.append(i)\n else:\n stack.pop()\n if len(st... | 7 | Given a string containing just the characters `'('` and `')'`, return _the length of the longest valid (well-formed) parentheses_ _substring_.
**Example 1:**
**Input:** s = "(() "
**Output:** 2
**Explanation:** The longest valid parentheses substring is "() ".
**Example 2:**
**Input:** s = ")()()) "
**Output:** ... | null |
Python || 97.04% Faster || Two Approaches || O(1) and o(n) Space | longest-valid-parentheses | 0 | 1 | **Brute Force O(n) Space:**\n```\nclass Solution:\n def longestValidParentheses(self, s: str) -> int:\n st=[-1]\n m=0\n for i in range(len(s)):\n if s[i]==\'(\':\n st.append(i)\n else:\n st.pop()\n if not st: #if -1 is popped\n ... | 5 | Given a string containing just the characters `'('` and `')'`, return _the length of the longest valid (well-formed) parentheses_ _substring_.
**Example 1:**
**Input:** s = "(() "
**Output:** 2
**Explanation:** The longest valid parentheses substring is "() ".
**Example 2:**
**Input:** s = ")()()) "
**Output:** ... | null |
✅ 100% Binary Search Easy [VIDEO] O(log(n)) Optimal Solution | search-in-rotated-sorted-array | 1 | 1 | # Problem Understanding\n\nThe task is to search for a target integer in a sorted array that has been rotated at an unknown pivot. \n\nFor instance, the array $$[0,1,2,4,5,6,7]$$ could be rotated at the 4th position to give $$[4,5,6,7,0,1,2]$$. The challenge is to find the position of the target integer in this rotated... | 262 | There is an integer array `nums` sorted in ascending order (with **distinct** values).
Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k` (`1 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` ... | null |
Python short and clean. Single Binary-Search 2 solutions. | search-in-rotated-sorted-array | 0 | 1 | # Approach 1\nTL;DR, Similar to [Editorial solution Approach 3](https://leetcode.com/problems/search-in-rotated-sorted-array/editorial/) but shorter and cleaner.\ni.e Compares subarray limits first and then the target.\n\n# Complexity\n- Time complexity: $$O(log_2(n))$$\n\n- Space complexity: $$O(1)$$\n\n# Code\n```pyt... | 1 | There is an integer array `nums` sorted in ascending order (with **distinct** values).
Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k` (`1 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` ... | null |
three lines of python | search-in-rotated-sorted-array | 0 | 1 | If I were writing real code, then for the sake of clarity I would (at the very least) expand the second line into an if-else block. But it\'s still fun to solve in as few lines as possible.\n\n# Complexity\n- Time complexity: O(log(n))\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1) a... | 1 | There is an integer array `nums` sorted in ascending order (with **distinct** values).
Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k` (`1 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` ... | null |
Python 3 || 7 lines, binary searches, w/ explanation || T/M: 92% / 94% | search-in-rotated-sorted-array | 0 | 1 | Here\'s the intuition:\n\n1. If we can determine`k`, the so-called *unknown pivot index*, the solution becomes easier. The key is that `nums[k]` is the first element in `nums` that is strictly less than `nums[0]`, so we can find it in *O*(log*N*) time with a boolean binary search.\n\n1. We now have the two subrrays, `n... | 5 | There is an integer array `nums` sorted in ascending order (with **distinct** values).
Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k` (`1 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` ... | null |
easy by Binary search.... | search-in-rotated-sorted-array | 0 | 1 | #first find the index of minimum number\n#split the array into two halfs\n#apply Binary search on both the arrays \n\nclass Solution:\n def search(self, nums: List[int], target: int) -> int:\n l=0\n l3=[]\n h=len(nums)-1\n mi=-1\n#Binary seaarch to first find the index of minimum number\n... | 1 | There is an integer array `nums` sorted in ascending order (with **distinct** values).
Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k` (`1 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` ... | null |
Easiest Solution | Binary Search | search-in-rotated-sorted-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(logn)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$... | 1 | There is an integer array `nums` sorted in ascending order (with **distinct** values).
Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k` (`1 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` ... | null |
C++ simple solution 15 lines binary search | search-in-rotated-sorted-array | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(log N)\n<!-- Add your space complexity here, e.g... | 1 | There is an integer array `nums` sorted in ascending order (with **distinct** values).
Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k` (`1 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` ... | null |
C++/Python binary search|| beats 100% | search-in-rotated-sorted-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis problem is solved by using binary search several times.\nFirstly, it is suggested to find the location for the peak which can be done by binary search.\nSecondly, using binary search to find location for the target which is possible ... | 6 | There is an integer array `nums` sorted in ascending order (with **distinct** values).
Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k` (`1 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` ... | null |
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