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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Simple Python3 Solution || 1 Line Code O(n) || Upto 98 % Space Saver O(1) | majority-element-ii | 0 | 1 | # Intuition\nWe use Counter to generate a Dictionary of nums and count of each distint number. Using this we filter the only the ones that are repeated more than n/3 times where n is the lenght of nums.\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution:\n def major... | 0 | Given an integer array of size `n`, find all elements that appear more than `⌊ n/3 ⌋` times.
**Example 1:**
**Input:** nums = \[3,2,3\]
**Output:** \[3\]
**Example 2:**
**Input:** nums = \[1\]
**Output:** \[1\]
**Example 3:**
**Input:** nums = \[1,2\]
**Output:** \[1,2\]
**Constraints:**
* `1 <= nums.length <... | How many majority elements could it possibly have?
Do you have a better hint? Suggest it! |
Easy +80% 9 line Python solution | majority-element-ii | 0 | 1 | \n# Code\n```py\nclass Solution:\n def majorityElement(self, nums: List[int]) -> List[int]:\n ans = []\n dct = {}\n n = len(nums)\n\n for x in set(nums):\n dct[x] = nums.count(x)\n\n for k, x in dct.items():\n if x > n//3:\n ans.append(k)\n\n ... | 1 | Given an integer array of size `n`, find all elements that appear more than `⌊ n/3 ⌋` times.
**Example 1:**
**Input:** nums = \[3,2,3\]
**Output:** \[3\]
**Example 2:**
**Input:** nums = \[1\]
**Output:** \[1\]
**Example 3:**
**Input:** nums = \[1,2\]
**Output:** \[1,2\]
**Constraints:**
* `1 <= nums.length <... | How many majority elements could it possibly have?
Do you have a better hint? Suggest it! |
Easy +80% 9 line Python solution | majority-element-ii | 0 | 1 | \n# Code\n```py\nclass Solution:\n def majorityElement(self, nums: List[int]) -> List[int]:\n ans = []\n dct = {}\n n = len(nums)\n\n for x in set(nums):\n dct[x] = nums.count(x)\n\n for k, x in dct.items():\n if x > n//3:\n ans.append(k)\n\n ... | 1 | Given an integer array of size `n`, find all elements that appear more than `⌊ n/3 ⌋` times.
**Example 1:**
**Input:** nums = \[3,2,3\]
**Output:** \[3\]
**Example 2:**
**Input:** nums = \[1\]
**Output:** \[1\]
**Example 3:**
**Input:** nums = \[1,2\]
**Output:** \[1,2\]
**Constraints:**
* `1 <= nums.length <... | How many majority elements could it possibly have?
Do you have a better hint? Suggest it! |
Hashtable and Count Method | majority-element-ii | 0 | 1 | # Hashtable Method:TC-->O(N)\n```\nclass Solution:\n def majorityElement(self, nums: List[int]) -> List[int]:\n dic=Counter(nums)\n n,list1=len(nums),[]\n for i,v in dic.items():\n if v>n//3:\n list1.append(i)\n return list1\n```\n# Count Method--->(N^2)\n```\ncl... | 5 | Given an integer array of size `n`, find all elements that appear more than `⌊ n/3 ⌋` times.
**Example 1:**
**Input:** nums = \[3,2,3\]
**Output:** \[3\]
**Example 2:**
**Input:** nums = \[1\]
**Output:** \[1\]
**Example 3:**
**Input:** nums = \[1,2\]
**Output:** \[1,2\]
**Constraints:**
* `1 <= nums.length <... | How many majority elements could it possibly have?
Do you have a better hint? Suggest it! |
Hashtable and Count Method | majority-element-ii | 0 | 1 | # Hashtable Method:TC-->O(N)\n```\nclass Solution:\n def majorityElement(self, nums: List[int]) -> List[int]:\n dic=Counter(nums)\n n,list1=len(nums),[]\n for i,v in dic.items():\n if v>n//3:\n list1.append(i)\n return list1\n```\n# Count Method--->(N^2)\n```\ncl... | 5 | Given an integer array of size `n`, find all elements that appear more than `⌊ n/3 ⌋` times.
**Example 1:**
**Input:** nums = \[3,2,3\]
**Output:** \[3\]
**Example 2:**
**Input:** nums = \[1\]
**Output:** \[1\]
**Example 3:**
**Input:** nums = \[1,2\]
**Output:** \[1,2\]
**Constraints:**
* `1 <= nums.length <... | How many majority elements could it possibly have?
Do you have a better hint? Suggest it! |
Most Simplest Approach | kth-smallest-element-in-a-bst | 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 the `root` of a binary search tree, and an integer `k`, return _the_ `kth` _smallest value (**1-indexed**) of all the values of the nodes in the tree_.
**Example 1:**
**Input:** root = \[3,1,4,null,2\], k = 1
**Output:** 1
**Example 2:**
**Input:** root = \[5,3,6,2,4,null,null,1\], k = 3
**Output:** 3
**Cons... | Try to utilize the property of a BST. Try in-order traversal. (Credits to @chan13) What if you could modify the BST node's structure? The optimal runtime complexity is O(height of BST). |
Recursive approach✅ | O( n)✅ | (Step by step explanation)✅ | kth-smallest-element-in-a-bst | 0 | 1 | # Intuition\nThe problem asks for finding the kth smallest element in a binary search tree (BST). A binary search tree is a binary tree where, for each node, the values of all nodes in its left subtree are less than the node\'s value, and the values of all nodes in its right subtree are greater than the node\'s value.\... | 3 | Given the `root` of a binary search tree, and an integer `k`, return _the_ `kth` _smallest value (**1-indexed**) of all the values of the nodes in the tree_.
**Example 1:**
**Input:** root = \[3,1,4,null,2\], k = 1
**Output:** 1
**Example 2:**
**Input:** root = \[5,3,6,2,4,null,null,1\], k = 3
**Output:** 3
**Cons... | Try to utilize the property of a BST. Try in-order traversal. (Credits to @chan13) What if you could modify the BST node's structure? The optimal runtime complexity is O(height of BST). |
Confused? Click this! Unlock the secrets of BSTs! | kth-smallest-element-in-a-bst | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe property of BST allows us to traverse it from smallest node to largest simply with an **in-order DFS**.\n\nWhat is the smallest node in a BST? It\'s the leftmost node. What\'s the second smallest? It\'s the leftmost node\'s parent, p.... | 5 | Given the `root` of a binary search tree, and an integer `k`, return _the_ `kth` _smallest value (**1-indexed**) of all the values of the nodes in the tree_.
**Example 1:**
**Input:** root = \[3,1,4,null,2\], k = 1
**Output:** 1
**Example 2:**
**Input:** root = \[5,3,6,2,4,null,null,1\], k = 3
**Output:** 3
**Cons... | Try to utilize the property of a BST. Try in-order traversal. (Credits to @chan13) What if you could modify the BST node's structure? The optimal runtime complexity is O(height of BST). |
✅🔥 Recursion & Vector Solutions with Complexity Analysis! - C++/Java/Python | kth-smallest-element-in-a-bst | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this solution is to perform an inorder traversal of the binary search tree to collect values in ascending order. Once the values are collected, the kth smallest value can be easily retrieved.\n\n# Approach 01\n<!-- De... | 105 | Given the `root` of a binary search tree, and an integer `k`, return _the_ `kth` _smallest value (**1-indexed**) of all the values of the nodes in the tree_.
**Example 1:**
**Input:** root = \[3,1,4,null,2\], k = 1
**Output:** 1
**Example 2:**
**Input:** root = \[5,3,6,2,4,null,null,1\], k = 3
**Output:** 3
**Cons... | Try to utilize the property of a BST. Try in-order traversal. (Credits to @chan13) What if you could modify the BST node's structure? The optimal runtime complexity is O(height of BST). |
Python3 (DFS/DFS Recursive/ BFS) | kth-smallest-element-in-a-bst | 0 | 1 | \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nDFS\nRecursive DFS\nBFS\n# Complexity\n- Time complexity: O(1)\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# Code\n```\n# Definition for a binary tre... | 2 | Given the `root` of a binary search tree, and an integer `k`, return _the_ `kth` _smallest value (**1-indexed**) of all the values of the nodes in the tree_.
**Example 1:**
**Input:** root = \[3,1,4,null,2\], k = 1
**Output:** 1
**Example 2:**
**Input:** root = \[5,3,6,2,4,null,null,1\], k = 3
**Output:** 3
**Cons... | Try to utilize the property of a BST. Try in-order traversal. (Credits to @chan13) What if you could modify the BST node's structure? The optimal runtime complexity is O(height of BST). |
Iterative approach✅ | O( n)✅ | (Step by step explanation)✅ | kth-smallest-element-in-a-bst | 0 | 1 | # Intuition\nThe problem requires finding the kth smallest element in a binary search tree (BST). A binary search tree is a binary tree where, for each node, the values of all nodes in its left subtree are less than the node\'s value, and the values of all nodes in its right subtree are greater than the node\'s value.\... | 3 | Given the `root` of a binary search tree, and an integer `k`, return _the_ `kth` _smallest value (**1-indexed**) of all the values of the nodes in the tree_.
**Example 1:**
**Input:** root = \[3,1,4,null,2\], k = 1
**Output:** 1
**Example 2:**
**Input:** root = \[5,3,6,2,4,null,null,1\], k = 3
**Output:** 3
**Cons... | Try to utilize the property of a BST. Try in-order traversal. (Credits to @chan13) What if you could modify the BST node's structure? The optimal runtime complexity is O(height of BST). |
|| Binary Search Tree + Heap(Priority Queue) || Easy Solution | kth-smallest-element-in-a-bst | 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 the `root` of a binary search tree, and an integer `k`, return _the_ `kth` _smallest value (**1-indexed**) of all the values of the nodes in the tree_.
**Example 1:**
**Input:** root = \[3,1,4,null,2\], k = 1
**Output:** 1
**Example 2:**
**Input:** root = \[5,3,6,2,4,null,null,1\], k = 3
**Output:** 3
**Cons... | Try to utilize the property of a BST. Try in-order traversal. (Credits to @chan13) What if you could modify the BST node's structure? The optimal runtime complexity is O(height of BST). |
3 different solutions, sort, recursive, stack | kth-smallest-element-in-a-bst | 0 | 1 | # Intuition\nUsing **sort** to grab the kth element\n\n# Complexity\n- Time complexity: O(n * log(n))\n- Space complexity: O(n)\n\n# Code\n```\ndef kthSmallest(self, root: Optional[TreeNode], k: int) -> int:\n res = []\n stack = [root]\n while stack:\n node = stack.pop()\n if not node:\n ... | 1 | Given the `root` of a binary search tree, and an integer `k`, return _the_ `kth` _smallest value (**1-indexed**) of all the values of the nodes in the tree_.
**Example 1:**
**Input:** root = \[3,1,4,null,2\], k = 1
**Output:** 1
**Example 2:**
**Input:** root = \[5,3,6,2,4,null,null,1\], k = 3
**Output:** 3
**Cons... | Try to utilize the property of a BST. Try in-order traversal. (Credits to @chan13) What if you could modify the BST node's structure? The optimal runtime complexity is O(height of BST). |
Kth Largest And Kth Smallest Element : C++ / Java / Python / Kotlin | kth-smallest-element-in-a-bst | 1 | 1 | # Intuition\nSo, the intuition is very straight forward, we need to traverse the given Binary Search Tree and find the kth **Smallest** and **Largest** element in it.\n\nThere are a couple of ways of doing this.\n\n**TIME & SPACE Complexity is discussed below in detail**\n\n<br>\n<hr>\n\n# Kth Smallest Element :\n- We ... | 2 | Given the `root` of a binary search tree, and an integer `k`, return _the_ `kth` _smallest value (**1-indexed**) of all the values of the nodes in the tree_.
**Example 1:**
**Input:** root = \[3,1,4,null,2\], k = 1
**Output:** 1
**Example 2:**
**Input:** root = \[5,3,6,2,4,null,null,1\], k = 3
**Output:** 3
**Cons... | Try to utilize the property of a BST. Try in-order traversal. (Credits to @chan13) What if you could modify the BST node's structure? The optimal runtime complexity is O(height of BST). |
Python3 - iterative dfs (inorder) beats 98% | kth-smallest-element-in-a-bst | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSince it\'s a BST traversing it in inorder way will give me a sorted list\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\ntraverse the nodes of tree in inorder way keep a count(initialise with 0) of the node and wh... | 1 | Given the `root` of a binary search tree, and an integer `k`, return _the_ `kth` _smallest value (**1-indexed**) of all the values of the nodes in the tree_.
**Example 1:**
**Input:** root = \[3,1,4,null,2\], k = 1
**Output:** 1
**Example 2:**
**Input:** root = \[5,3,6,2,4,null,null,1\], k = 3
**Output:** 3
**Cons... | Try to utilize the property of a BST. Try in-order traversal. (Credits to @chan13) What if you could modify the BST node's structure? The optimal runtime complexity is O(height of BST). |
230: Time 97.61% and Space 83.27%, Solution with step by step explanation | kth-smallest-element-in-a-bst | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThe problem asks us to find the kth smallest value in a binary search tree. One way to do this is to traverse the tree in-order (left, root, right) and keep track of the number of nodes we have visited. Once we have visited ... | 14 | Given the `root` of a binary search tree, and an integer `k`, return _the_ `kth` _smallest value (**1-indexed**) of all the values of the nodes in the tree_.
**Example 1:**
**Input:** root = \[3,1,4,null,2\], k = 1
**Output:** 1
**Example 2:**
**Input:** root = \[5,3,6,2,4,null,null,1\], k = 3
**Output:** 3
**Cons... | Try to utilize the property of a BST. Try in-order traversal. (Credits to @chan13) What if you could modify the BST node's structure? The optimal runtime complexity is O(height of BST). |
Python || Recursion || Without Stack || Easy | kth-smallest-element-in-a-bst | 0 | 1 | ```\nclass Solution:\n def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:\n def inorder(root):\n nonlocal k\n if root==None:\n return None\n left=inorder(root.left)\n if left!=None:\n return left\n k-=1\n ... | 4 | Given the `root` of a binary search tree, and an integer `k`, return _the_ `kth` _smallest value (**1-indexed**) of all the values of the nodes in the tree_.
**Example 1:**
**Input:** root = \[3,1,4,null,2\], k = 1
**Output:** 1
**Example 2:**
**Input:** root = \[5,3,6,2,4,null,null,1\], k = 3
**Output:** 3
**Cons... | Try to utilize the property of a BST. Try in-order traversal. (Credits to @chan13) What if you could modify the BST node's structure? The optimal runtime complexity is O(height of BST). |
Simple Solution Inorder Traversal | DFS | Beats 93% Runtime | Python | kth-smallest-element-in-a-bst | 0 | 1 | \n# Code\n```\n# Definition for a binary tree node.\n# class TreeNode:\n# def __init__(self, val=0, left=None, right=None):\n# self.val = val\n# self.left = left\n# self.right = right\nclass Solution:\n def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:\n\n preorder = ... | 3 | Given the `root` of a binary search tree, and an integer `k`, return _the_ `kth` _smallest value (**1-indexed**) of all the values of the nodes in the tree_.
**Example 1:**
**Input:** root = \[3,1,4,null,2\], k = 1
**Output:** 1
**Example 2:**
**Input:** root = \[5,3,6,2,4,null,null,1\], k = 3
**Output:** 3
**Cons... | Try to utilize the property of a BST. Try in-order traversal. (Credits to @chan13) What if you could modify the BST node's structure? The optimal runtime complexity is O(height of BST). |
Beats 99% of other solutions ! | power-of-two | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def isPowerOfTwo(self, n: int) -> bool:\n # y = b^x -> x = logb y\n if n <=0:\n return False\n return math.log2(n) % 1 == 0 \n\n \n``` | 0 | Given an integer `n`, return _`true` if it is a power of two. Otherwise, return `false`_.
An integer `n` is a power of two, if there exists an integer `x` such that `n == 2x`.
**Example 1:**
**Input:** n = 1
**Output:** true
**Explanation:** 20 = 1
**Example 2:**
**Input:** n = 16
**Output:** true
**Explanation:**... | null |
🔥Simple 1 Line using Binary | power-of-two | 0 | 1 | \n# Code\n```\nclass Solution:\n def isPowerOfTwo(self, n: int) -> bool:\n return bin(n)[2] == "1" and bin(n).count("1") == 1\n```\n# Explanation\n\n**Every power of two `(2^n)` has a single `1` bit at the left side, and all the other bits are `0`.**\n\n*Here are some examples to illustrate this:*\n\n- 2^0 = ... | 5 | Given an integer `n`, return _`true` if it is a power of two. Otherwise, return `false`_.
An integer `n` is a power of two, if there exists an integer `x` such that `n == 2x`.
**Example 1:**
**Input:** n = 1
**Output:** true
**Explanation:** 20 = 1
**Example 2:**
**Input:** n = 16
**Output:** true
**Explanation:**... | null |
Beats 100% | Easy To Understand | 0ms 🔥🚀 | power-of-two | 1 | 1 | # Intuition\nPLease upvote if you found this helpful!!\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(1)\n\n\n- Space comp... | 2 | Given an integer `n`, return _`true` if it is a power of two. Otherwise, return `false`_.
An integer `n` is a power of two, if there exists an integer `x` such that `n == 2x`.
**Example 1:**
**Input:** n = 1
**Output:** true
**Explanation:** 20 = 1
**Example 2:**
**Input:** n = 16
**Output:** true
**Explanation:**... | null |
Very Easy 0 ms 100% (Fully Explained)(Java, C++, Python, JS, C, Python3) | power-of-two | 1 | 1 | # **Java Solution (Recursive Approach):**\nRuntime: 0 ms, faster than 100.00% of Java online submissions for Power of Two.\n```\nclass Solution {\n public boolean isPowerOfTwo(int n) {\n // Base cases as \'1\' is the only odd number which is a power of 2 i.e. 2^0...\n if(n==1)\n return true;... | 55 | Given an integer `n`, return _`true` if it is a power of two. Otherwise, return `false`_.
An integer `n` is a power of two, if there exists an integer `x` such that `n == 2x`.
**Example 1:**
**Input:** n = 1
**Output:** true
**Explanation:** 20 = 1
**Example 2:**
**Input:** n = 16
**Output:** true
**Explanation:**... | null |
linear solution - Python 🐍 | power-of-two | 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 `n`, return _`true` if it is a power of two. Otherwise, return `false`_.
An integer `n` is a power of two, if there exists an integer `x` such that `n == 2x`.
**Example 1:**
**Input:** n = 1
**Output:** true
**Explanation:** 20 = 1
**Example 2:**
**Input:** n = 16
**Output:** true
**Explanation:**... | null |
Easy Python Solution 🐍 | power-of-two | 0 | 1 | # Code\n```\nclass Solution:\n def isPowerOfTwo(self, n: int) -> bool:\n if n==1: return True\n x=2\n while x<=n:\n if x==n: return True\n x*=2\n return False\n```\n***Hope it helps...!!*** \uD83D\uDE07\u270C\uFE0F | 1 | Given an integer `n`, return _`true` if it is a power of two. Otherwise, return `false`_.
An integer `n` is a power of two, if there exists an integer `x` such that `n == 2x`.
**Example 1:**
**Input:** n = 1
**Output:** true
**Explanation:** 20 = 1
**Example 2:**
**Input:** n = 16
**Output:** true
**Explanation:**... | null |
One Line Solution using Bitwise Operator | power-of-two | 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(1)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $... | 1 | Given an integer `n`, return _`true` if it is a power of two. Otherwise, return `false`_.
An integer `n` is a power of two, if there exists an integer `x` such that `n == 2x`.
**Example 1:**
**Input:** n = 1
**Output:** true
**Explanation:** 20 = 1
**Example 2:**
**Input:** n = 16
**Output:** true
**Explanation:**... | null |
[Python] Using BitManipulation - Time Complex: O(1); Space Complex: O(1) | power-of-two | 0 | 1 | \n\n# Approach\nUsing BitManipulation\n\n# Complexity\n- Time complexity: O(32)\n\n- Space complexity: O(1)\n\n# Solution 1\nIf a number is power of two, the number of bit 1 in it must be one\n```\nclass Solution:\n def isPowerOfTwo(self, n: int) -> bool:\n count = 0\n # count is variable to calculate ... | 2 | Given an integer `n`, return _`true` if it is a power of two. Otherwise, return `false`_.
An integer `n` is a power of two, if there exists an integer `x` such that `n == 2x`.
**Example 1:**
**Input:** n = 1
**Output:** true
**Explanation:** 20 = 1
**Example 2:**
**Input:** n = 16
**Output:** true
**Explanation:**... | null |
Bit Minuplation: Beats 99.9% submissions | power-of-two | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe first thought is just by keep dividing the number by 2 until we cannot and check if the number is 1. If the number at the last is 1 return True otherwise false. This takes O(logn) time complexity.\n\n# Approach\n<!-- Describe your app... | 9 | Given an integer `n`, return _`true` if it is a power of two. Otherwise, return `false`_.
An integer `n` is a power of two, if there exists an integer `x` such that `n == 2x`.
**Example 1:**
**Input:** n = 1
**Output:** true
**Explanation:** 20 = 1
**Example 2:**
**Input:** n = 16
**Output:** true
**Explanation:**... | null |
Very Easy || 100% || Fully Explained || Java, C++, Python, Python3 | implement-queue-using-stacks | 1 | 1 | **Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).**\n\n**Implement the MyQueue class:**\n\n* void push(int x) Pushes element x to the back of the queue.\n* int pop() Removes the element from the f... | 190 | Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (`push`, `peek`, `pop`, and `empty`).
Implement the `MyQueue` class:
* `void push(int x)` Pushes element x to the back of the queue.
* `int pop()` Removes the element from th... | null |
232: Solution with step by step explanation | implement-queue-using-stacks | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThe key to implementing a queue using two stacks is to use one stack to represent the front of the queue and the other stack to represent the back of the queue. When an element is pushed onto the queue, it is pushed onto the... | 12 | Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (`push`, `peek`, `pop`, and `empty`).
Implement the `MyQueue` class:
* `void push(int x)` Pushes element x to the back of the queue.
* `int pop()` Removes the element from th... | null |
very easyy to understand | implement-queue-using-stacks | 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 | Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (`push`, `peek`, `pop`, and `empty`).
Implement the `MyQueue` class:
* `void push(int x)` Pushes element x to the back of the queue.
* `int pop()` Removes the element from th... | null |
Solution | number-of-digit-one | 1 | 1 | ```C++ []\nclass Solution { \npublic:\n int countDigitOne(int n) {\n int ret = 0;\n for(long long int i = 1; i <= n; i*= (long long int)10){\n int a = n / i;\n int b = n % i;\n int x = a % 10;\n if(x ==1){\n ret += (a / 10) * i + (b + 1);\n }\n else ... | 430 | Given an integer `n`, count _the total number of digit_ `1` _appearing in all non-negative integers less than or equal to_ `n`.
**Example 1:**
**Input:** n = 13
**Output:** 6
**Example 2:**
**Input:** n = 0
**Output:** 0
**Constraints:**
* `0 <= n <= 109` | Beware of overflow. |
Python 3 solution || 100% working 🔥🔥🔥 | number-of-digit-one | 0 | 1 | ****Please Upvote :-)**** \n\n# Complexity\n- Time complexity: O(logn)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->Since it loops only the number of digits present in a number.\n\n# Code\n```\nclass Solution:\n def countDigitOne(self, n: int) :\n n = str(n) #converting it to string\n ... | 2 | Given an integer `n`, count _the total number of digit_ `1` _appearing in all non-negative integers less than or equal to_ `n`.
**Example 1:**
**Input:** n = 13
**Output:** 6
**Example 2:**
**Input:** n = 0
**Output:** 0
**Constraints:**
* `0 <= n <= 109` | Beware of overflow. |
233: Solution with step by step explanation | number-of-digit-one | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThe solution works by iterating through each digit position from right to left, and for each position, counting the number of ones that can appear at that position. This is done using the formula:\n\n- If the digit at the cu... | 11 | Given an integer `n`, count _the total number of digit_ `1` _appearing in all non-negative integers less than or equal to_ `n`.
**Example 1:**
**Input:** n = 13
**Output:** 6
**Example 2:**
**Input:** n = 0
**Output:** 0
**Constraints:**
* `0 <= n <= 109` | Beware of overflow. |
Easy Recursive Python Solution | number-of-digit-one | 0 | 1 | # Intuition\nAs someone pointed out in the comments of the official solution, you can avoid the complicated math by using a formula for the sum of ones in numbers **strictly less** than 10, 100, 1000, etc.\n\nTake 1000 for example. The numbers strictly less than 1000 are 000-999. Because the digits that make up these n... | 3 | Given an integer `n`, count _the total number of digit_ `1` _appearing in all non-negative integers less than or equal to_ `n`.
**Example 1:**
**Input:** n = 13
**Output:** 6
**Example 2:**
**Input:** n = 0
**Output:** 0
**Constraints:**
* `0 <= n <= 109` | Beware of overflow. |
Explained to a 5-year old | number-of-digit-one | 0 | 1 | # Intuition\nThe key intuition is that we count the number of 1\'s contributed by every digit\'s place in the number.\nFor example, if `n` is a 3-digit number, some of the 1\'s would be contributed by the 1\'s place, some by the 10\'s place and some by the 100\'s place.\n\n# Approach\nObserve that for every range of 10... | 1 | Given an integer `n`, count _the total number of digit_ `1` _appearing in all non-negative integers less than or equal to_ `n`.
**Example 1:**
**Input:** n = 13
**Output:** 6
**Example 2:**
**Input:** n = 0
**Output:** 0
**Constraints:**
* `0 <= n <= 109` | Beware of overflow. |
10 lines python code, beats 92% of people | number-of-digit-one | 0 | 1 | # Intuition\neach digit of a number has a possibility of being a one, and so we can count them up using a little math and some loops\n\n# Approach\nMultiplier represents the place value, and divider serves as to remove anything above that place value. Then two additions to count:\nOne for the number of ones you use for... | 0 | Given an integer `n`, count _the total number of digit_ `1` _appearing in all non-negative integers less than or equal to_ `n`.
**Example 1:**
**Input:** n = 13
**Output:** 6
**Example 2:**
**Input:** n = 0
**Output:** 0
**Constraints:**
* `0 <= n <= 109` | Beware of overflow. |
Easy Solution : ) | number-of-digit-one | 1 | 1 | # Intuition\nThe problem requires counting the number of occurrences of the digit \'1\' in all non-negative integers less than or equal to a given integer n. To approach this problem, we can iterate through each digit place of the numbers, count the occurrences of \'1\' at that place, and sum them up to get the total c... | 0 | Given an integer `n`, count _the total number of digit_ `1` _appearing in all non-negative integers less than or equal to_ `n`.
**Example 1:**
**Input:** n = 13
**Output:** 6
**Example 2:**
**Input:** n = 0
**Output:** 0
**Constraints:**
* `0 <= n <= 109` | Beware of overflow. |
pure math solution | number-of-digit-one | 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 an integer `n`, count _the total number of digit_ `1` _appearing in all non-negative integers less than or equal to_ `n`.
**Example 1:**
**Input:** n = 13
**Output:** 6
**Example 2:**
**Input:** n = 0
**Output:** 0
**Constraints:**
* `0 <= n <= 109` | Beware of overflow. |
Fast Memoization Python | number-of-digit-one | 0 | 1 | class Solution:\n def countDigitOne(self, n: int) -> int:\n @cache\n def f(k):\n if k == 0:\n return 0\n if k <= 9:\n return 1\n\n s = str(k)\n head_num = int(s[0])\n rest_num = int(s[1:])\n\n return f(rest_... | 0 | Given an integer `n`, count _the total number of digit_ `1` _appearing in all non-negative integers less than or equal to_ `n`.
**Example 1:**
**Input:** n = 13
**Output:** 6
**Example 2:**
**Input:** n = 0
**Output:** 0
**Constraints:**
* `0 <= n <= 109` | Beware of overflow. |
19ms Beats 100.00% of users with Python3 | T/S: O((log(n))^2)/O(log(n)) | number-of-digit-one | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe idea used here is to count the number of times 1 can appear in each digits position such that the number is less than or equal to n. Once you have the counts for each digits place, just add them.\nExample:\nlets say n=130,\nNumber of ... | 0 | Given an integer `n`, count _the total number of digit_ `1` _appearing in all non-negative integers less than or equal to_ `n`.
**Example 1:**
**Input:** n = 13
**Output:** 6
**Example 2:**
**Input:** n = 0
**Output:** 0
**Constraints:**
* `0 <= n <= 109` | Beware of overflow. |
Python Solution | number-of-digit-one | 0 | 1 | # Complexity\n- Time complexity:\nO(log 10(n))\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution:\n def countDigitOne(self, n: int) -> int:\n cnt = 0\n i = 1\n while i <= n:\n div = i * 10\n cnt += (n//div) * i + min(max(n % div - i + 1, 0), i)\n i*=10\... | 0 | Given an integer `n`, count _the total number of digit_ `1` _appearing in all non-negative integers less than or equal to_ `n`.
**Example 1:**
**Input:** n = 13
**Output:** 6
**Example 2:**
**Input:** n = 0
**Output:** 0
**Constraints:**
* `0 <= n <= 109` | Beware of overflow. |
Mathematical approach; Explained w/ clean code | number-of-digit-one | 0 | 1 | # Intuition\nThis approach counts the number of digit 1s in non-negative integers less than or equal to n. \n\n# Approach\nThe code iterates through the digits of n and calculates the count of digit 1s at each position. It uses the `before`, `curr`, and `after` variables to determine the count efficiently.\n\n# Complex... | 0 | Given an integer `n`, count _the total number of digit_ `1` _appearing in all non-negative integers less than or equal to_ `n`.
**Example 1:**
**Input:** n = 13
**Output:** 6
**Example 2:**
**Input:** n = 0
**Output:** 0
**Constraints:**
* `0 <= n <= 109` | Beware of overflow. |
[Explanation] Dynamic Programming | number-of-digit-one | 0 | 1 | Given an integer n, the task is to determine the total number of occurrences of the digit 1 in all integers from 0 to n, inclusive.\n\n##### Approach: Dynamic Programming\n\nUnderstanding the concept involves breaking down the numbers into their digits and calculating the count of 1\'s at each digit place. Dynamic prog... | 0 | Given an integer `n`, count _the total number of digit_ `1` _appearing in all non-negative integers less than or equal to_ `n`.
**Example 1:**
**Input:** n = 13
**Output:** 6
**Example 2:**
**Input:** n = 0
**Output:** 0
**Constraints:**
* `0 <= n <= 109` | Beware of overflow. |
No DP, Space O(1) | number-of-digit-one | 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(log)\n\n- Space complexity: O(1)\n\n# Code\n```\nclass Solution:\n def countDigitOne(self, n: int) -> int:\n \n s... | 0 | Given an integer `n`, count _the total number of digit_ `1` _appearing in all non-negative integers less than or equal to_ `n`.
**Example 1:**
**Input:** n = 13
**Output:** 6
**Example 2:**
**Input:** n = 0
**Output:** 0
**Constraints:**
* `0 <= n <= 109` | Beware of overflow. |
🔥🔥 3 Best Solutions Explained 🔥🔥 | palindrome-linked-list | 0 | 1 | https://youtu.be/8h2UOPDCGPM | 50 | Given the `head` of a singly linked list, return `true` _if it is a_ _palindrome_ _or_ `false` _otherwise_.
**Example 1:**
**Input:** head = \[1,2,2,1\]
**Output:** true
**Example 2:**
**Input:** head = \[1,2\]
**Output:** false
**Constraints:**
* The number of nodes in the list is in the range `[1, 105]`.
* ... | null |
Python | Recursion | palindrome-linked-list | 0 | 1 | # Code\n```\nclass Solution:\n def isPalindrome(self, head: Optional[ListNode]) -> bool:\n if not head or not head.next: return True\n fp = sp = head\n while sp: \n fp, sp = fp.next, sp.next\n if sp: sp = sp.next\n return self.helper(fp, head)[0]\n \n def helpe... | 1 | Given the `head` of a singly linked list, return `true` _if it is a_ _palindrome_ _or_ `false` _otherwise_.
**Example 1:**
**Input:** head = \[1,2,2,1\]
**Output:** true
**Example 2:**
**Input:** head = \[1,2\]
**Output:** false
**Constraints:**
* The number of nodes in the list is in the range `[1, 105]`.
* ... | null |
Easy and Fundamental Solution For Linked Lists (In-Depth explanation to fundamentals (beats 89%)) | palindrome-linked-list | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nI wanted to solve this using the fundamentals of linked lists. So I split it up into 3 parts, using differnt strategies to do each step. This problem is a combination of ones like "Reverse Linked List" and "Middle of the Linked List"\n# A... | 36 | Given the `head` of a singly linked list, return `true` _if it is a_ _palindrome_ _or_ `false` _otherwise_.
**Example 1:**
**Input:** head = \[1,2,2,1\]
**Output:** true
**Example 2:**
**Input:** head = \[1,2\]
**Output:** false
**Constraints:**
* The number of nodes in the list is in the range `[1, 105]`.
* ... | null |
Beats 99%✅ | O( log(n))✅ | (Step by step explanation)✅ | lowest-common-ancestor-of-a-binary-search-tree | 0 | 1 | # Intuition\nThe problem is to find the lowest common ancestor (LCA) of two nodes, `p` and `q`, in a binary search tree (BST). The LCA is the lowest node in the tree that has both `p` and `q` as descendants. Since the BST property holds, we can efficiently traverse the tree to find the LCA.\n\n# Approach\nThe approach ... | 8 | Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has bo... | null |
Easy Recursive Solution | lowest-common-ancestor-of-a-binary-search-tree | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo solve the problem, we can traverse the tree recursively, searching for both target nodes in the left and right subtrees. If both target nodes are found in different subtrees, then the current node is the lowest common ancestor. If they... | 2 | Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has bo... | null |
Recursive approach✅ | O( log(n))✅ | C++(Step by step explanation)✅ | lowest-common-ancestor-of-a-binary-search-tree | 0 | 1 | # Intuition\nThe problem is to find the lowest common ancestor (LCA) of two nodes, `p` and `q`, in a binary search tree (BST). The LCA is the lowest node in the tree that has both `p` and `q` as descendants. Since the BST property holds, we can efficiently traverse the tree to find the LCA.\n\n# Approach\nThe approach ... | 5 | Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has bo... | null |
Explained Easy Iterative Python Solution | lowest-common-ancestor-of-a-binary-search-tree | 0 | 1 | We will rely upon the invariant of the BST to solve the exercise. We know that the left subtree of each node contains nodes with smaller values and the right subtree contains nodes with greater values. We also know that if two nodes, x and y, are on different subtrees of a node z (one in the left portion and one in the... | 103 | Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has bo... | null |
Simple solution in two way python | lowest-common-ancestor-of-a-binary-search-tree | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nGiven a binary search tree (BST) and two nodes, we need to find the lowest common ancestor (LCA) node of both nodes in the BST. A LCA is the lowest node in the tree that has both nodes as its descendants.\n# Approach\n<!-- Describe your a... | 3 | Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has bo... | null |
Python; easy; runtime 95.94% | lowest-common-ancestor-of-a-binary-search-tree | 0 | 1 | \n\ncase i)\nif root.val is the smallest or the greatest among p.val, q.val, and root.val, then traverse down\n\ncase ii)\nif case (i) is false <=>\nif root.val is in between p.val and q.val: p and q are sepera... | 12 | Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has bo... | null |
235: Space 96.79%, Solution with step by step explanation | lowest-common-ancestor-of-a-binary-search-tree | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. We are given the root of a binary search tree and two nodes p and q.\n2. We start traversing from the root of the binary search tree.\n3. If the value of both p and q is less than the value of the root, then the LCA will ... | 17 | Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has bo... | null |
Using recursion | lowest-common-ancestor-of-a-binary-search-tree | 0 | 1 | # Intuition\nif p and q are on opposite sides of nodes, then the root will be the common node.\n\n# Approach\nIf they are on the same side, do recursion on root.left or root.right\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space compl... | 1 | Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has bo... | null |
[python] Easy Solution | lowest-common-ancestor-of-a-binary-search-tree | 0 | 1 | \n\n# Code\n```\n# Definition for a binary tree node.\n# class TreeNode:\n# def __init__(self, x):\n# self.val = x\n# self.left = None\n# self.right = None\n\nclass Solution:\n def lowestCommonAncestor(self, root: \'TreeNode\', p: \'TreeNode\', q: \'TreeNode\') -> \'TreeNode\':\n c... | 6 | Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has bo... | null |
Easy Iterative solution using traversal in Python. | lowest-common-ancestor-of-a-binary-search-tree | 0 | 1 | # Code\n```\nclass Solution:\n def lowestCommonAncestor(self, root: \'TreeNode\', p: \'TreeNode\', q: \'TreeNode\') -> \'TreeNode\':\n pv=p.val\n qv=q.val\n while(root!=None):\n if pv <root.val and qv<root.val:\n root=root.left\n elif pv>root.val and qv>root.... | 4 | Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has bo... | null |
✔️ Python Easy Way To Find Lowest Common Ancestor of a Binary Search Tree | 96% Faster 🔥 | lowest-common-ancestor-of-a-binary-search-tree | 0 | 1 | **IF YOU FIND THIS POST HELPFUL PLEASE UPVOTE \uD83D\uDC4D**\n\n```\nclass Solution:\n def lowestCommonAncestor(self, root: \'TreeNode\', p: \'TreeNode\', q: \'TreeNode\') -> \'TreeNode\':\n curr = root\n \n while curr:\n if p.val>curr.val and q.val>curr.val:\n curr = c... | 6 | Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has bo... | null |
Simple iterative solution, O(1) space, faster than 99.67% | lowest-common-ancestor-of-a-binary-search-tree | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAs this tree is a BST, it simplifies the problem a lot. Think about what property should hold true for a node to be the lowest common ancestor.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nOn thinking a bit, you... | 2 | Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has bo... | null |
236: Solution with step by step explanation | lowest-common-ancestor-of-a-binary-tree | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. The function lowestCommonAncestor takes in three parameters: the root of a binary tree (root) and two nodes of the binary tree (p and q).\n\n2. The first if statement checks if the root is None or if it is equal to either... | 81 | Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as... | null |
JAVA!! PYTHON!! Fastest Solution using Inorder Traversal | lowest-common-ancestor-of-a-binary-tree | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFind p and q. Whichever is found first can be the LCA if they are both on the same side. If they are not on the same side then LCA is root of the subtree.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nJust use no... | 0 | Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as... | null |
Easy Iterative DFS Solution in Python. | lowest-common-ancestor-of-a-binary-tree | 0 | 1 | # Code\n```\nclass Solution:\n def lowestCommonAncestor(self, root: \'TreeNode\', p: \'TreeNode\', q: \'TreeNode\') -> \'TreeNode\':\n stack=[]\n parent=defaultdict(TreeNode)\n res=[]\n stack.append([root,0])\n while(stack):\n t=stack.pop()\n node=t[0]\n ... | 4 | Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as... | null |
Easy, simple and most efficient recursive solution with explanation | lowest-common-ancestor-of-a-binary-tree | 1 | 1 | \n# Approach\n- Base Case: If the current node is null or is equal to either p or q, return the current node.\n\n- Recursive Calls: Recur on the left and right subtrees. These recursive calls return the LCA for p and q in their respective subtrees.\n\n- Combining Results: If both the left and right subtrees return non-... | 5 | Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as... | null |
Python 3 O(n) solution with explanation | lowest-common-ancestor-of-a-binary-tree | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThere are 3 possible cases:\n 1. p and q on opposite side of tree\n 2. p desc of q\n 3. q desc of p\n# Approach\n<!-- Describe your approach to solving the problem. -->\nOnce we have found either p or q, search the ot... | 5 | Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as... | null |
Python || Easy || O(n) ||Recursive Solution | lowest-common-ancestor-of-a-binary-tree | 0 | 1 | ```\nclass Solution:\n def lowestCommonAncestor(self, root: \'TreeNode\', p: \'TreeNode\', q: \'TreeNode\') -> \'TreeNode\':\n if root==None or root==p or root==q:\n return root\n left=self.lowestCommonAncestor(root.left,p,q)\n right=self.lowestCommonAncestor(root.right,p,q)\n ... | 8 | Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): "The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as... | null |
Easy and beginner friendly solution with Time complexity O(1) and Space complexity O(1) | delete-node-in-a-linked-list | 1 | 1 | # Intuition\n\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem provides a node to be deleted in a linked list, emphasizing the absence of duplicate values. Our strategy involves an in-place modification of the linked list by simply skipping the designated node.\n\n# Approach\n\n<!-- Des... | 1 | There is a singly-linked list `head` and we want to delete a node `node` in it.
You are given the node to be deleted `node`. You will **not be given access** to the first node of `head`.
All the values of the linked list are **unique**, and it is guaranteed that the given node `node` is not the last node in the linke... | null |
Simple and easy 1 min explanation in Python and JAVA | product-of-array-except-self | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSimilar to finding Prefix Sum array, here the preblem intends us to find the Prefix Product Array and Suffix Product Array for our original array.\n```\npre[i+1] = pre[i] * a[i]\nsuff[i-1] = suff[i] * a[i]\n```\n\n# Complexity\n- Time com... | 99 | Given an integer array `nums`, return _an array_ `answer` _such that_ `answer[i]` _is equal to the product of all the elements of_ `nums` _except_ `nums[i]`.
The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.
You must write an algorithm that runs in `O(n)` time and without... | null |
[VIDEO] Visualization and Explanation of O(n) Solution | product-of-array-except-self | 0 | 1 | https://youtu.be/5bS636lE_R0?si=CXD7IT3sxRkONnAe\n\nThe brute force method would be to manually calculate each product. Since there are n products to calculate, and each product is made up of n-1 numbers, this runs in O(n<sup>2</sup>) time.\n\nThis method causes us to do the same calculations over and over again. So ... | 101 | Given an integer array `nums`, return _an array_ `answer` _such that_ `answer[i]` _is equal to the product of all the elements of_ `nums` _except_ `nums[i]`.
The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.
You must write an algorithm that runs in `O(n)` time and without... | null |
✅ O(N) Single For-Loop Solution in Python (Easy to Understand) | product-of-array-except-self | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nLike any other problem, my first intuition was let\'s brute force the solution by having two pointers going from i-1 to 1 and i+1 to len(nums) and multiplying each element and then appending this result to the output array. However, on se... | 185 | Given an integer array `nums`, return _an array_ `answer` _such that_ `answer[i]` _is equal to the product of all the elements of_ `nums` _except_ `nums[i]`.
The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.
You must write an algorithm that runs in `O(n)` time and without... | null |
Easy and Simple Approach | product-of-array-except-self | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSimple Solution in python\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nDivide the conditions based on zeroes\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n)\n\n- ... | 0 | Given an integer array `nums`, return _an array_ `answer` _such that_ `answer[i]` _is equal to the product of all the elements of_ `nums` _except_ `nums[i]`.
The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.
You must write an algorithm that runs in `O(n)` time and without... | null |
C++ & Python 3 || in-place method || time: O(n) space: O(1) | product-of-array-except-self | 0 | 1 | # Intuition\nUsing division operation, and in-palce method.\nThere are 3 situations that need to be handled here.\n> 1. There is 1 `0` in arr.\n> 2. There are 2 and more `0` in arr.\n> 3. There is no `0` in arr.\n\n# Approach\n`single_zero` represent there is only one 0 in arr.\n`muti_zero` represent there are multiple... | 2 | Given an integer array `nums`, return _an array_ `answer` _such that_ `answer[i]` _is equal to the product of all the elements of_ `nums` _except_ `nums[i]`.
The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.
You must write an algorithm that runs in `O(n)` time and without... | null |
238: Time 96.95%, Solution with step by step explanation | product-of-array-except-self | 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:\n96.95%\n\n- Space complexity:\n68.31%\n\n# Code\n```\nclass Solution:\n def productExceptSelf(self, nums: List[int]) -> List[int]:... | 48 | Given an integer array `nums`, return _an array_ `answer` _such that_ `answer[i]` _is equal to the product of all the elements of_ `nums` _except_ `nums[i]`.
The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.
You must write an algorithm that runs in `O(n)` time and without... | null |
Product of Array Except Self: So Easy, Even a Caveman Can Do It! (Using Division Operation) | product-of-array-except-self | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this solution is to use the product of all the elements in the array, except for the current element, to calculate the answer for the current element. This can be done by maintaining a variable ```mul``` that keeps tr... | 2 | Given an integer array `nums`, return _an array_ `answer` _such that_ `answer[i]` _is equal to the product of all the elements of_ `nums` _except_ `nums[i]`.
The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.
You must write an algorithm that runs in `O(n)` time and without... | null |
Simple Python Solution | O(N) Time complexity | product-of-array-except-self | 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`, return _an array_ `answer` _such that_ `answer[i]` _is equal to the product of all the elements of_ `nums` _except_ `nums[i]`.
The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.
You must write an algorithm that runs in `O(n)` time and without... | null |
Simple Solution with Full Explanation || Easy to Follow || For Beginners | product-of-array-except-self | 0 | 1 | # Understanding Our Approach with an Example\n\nTo fully understand our approach it is best to follow along with an example.\n\nGiven a `nums` list **[1, 2, 3, 4]**, our algorithm works by first calculating each elements "left product." \n\nA list element\'s left product, in this case, is defined as the cumalative prod... | 3 | Given an integer array `nums`, return _an array_ `answer` _such that_ `answer[i]` _is equal to the product of all the elements of_ `nums` _except_ `nums[i]`.
The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.
You must write an algorithm that runs in `O(n)` time and without... | null |
Python🔥Java🔥C++🔥Simple Solution | sliding-window-maximum | 1 | 1 | # Video Solution \n\n# Search \uD83D\uDC49` Sliding Window Maximum By ErraK`\n\n# or \n\n# Click the Link in my Profile\n\n# An UPVOTE will be encouraging \uD83D\uDC4D\n\n```Python []\nclass Solution:\n def maxSlidingWindow(self, nums, k):\n result = []\n window = deque()\n\n for i, num in enume... | 27 | You are given an array of integers `nums`, there is a sliding window of size `k` which is moving from the very left of the array to the very right. You can only see the `k` numbers in the window. Each time the sliding window moves right by one position.
Return _the max sliding window_.
**Example 1:**
**Input:** nums... | How about using a data structure such as deque (double-ended queue)? The queue size need not be the same as the window’s size. Remove redundant elements and the queue should store only elements that need to be considered. |
Python3 Solution | sliding-window-maximum | 0 | 1 | \n```\nclass Solution:\n def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:\n n=len(nums)\n seen=[]\n ans=[] \n for i in range(n):\n if seen and seen[0]==i-k:\n seen.pop(0)\n\n while seen and nums[seen[-1]]<nums[i]:\n se... | 4 | You are given an array of integers `nums`, there is a sliding window of size `k` which is moving from the very left of the array to the very right. You can only see the `k` numbers in the window. Each time the sliding window moves right by one position.
Return _the max sliding window_.
**Example 1:**
**Input:** nums... | How about using a data structure such as deque (double-ended queue)? The queue size need not be the same as the window’s size. Remove redundant elements and the queue should store only elements that need to be considered. |
【Video】Ex-Amazon explains a solution with Python, JavaScript, Java and C++ | sliding-window-maximum | 1 | 1 | # Intuition\nUsing two pointers\n\n---\n\n# Solution Video\n\n### Please subscribe to my channel from here. I have 245 videos as of August 16th\nhttp://www.youtube.com/channel/UC9RMNwYTL3SXCP6ShLWVFww?sub_confirmation=1\n\nhttps://youtu.be/VYfy0VGa0_0\n\n### In the video, the steps of approach below are visualized usin... | 29 | You are given an array of integers `nums`, there is a sliding window of size `k` which is moving from the very left of the array to the very right. You can only see the `k` numbers in the window. Each time the sliding window moves right by one position.
Return _the max sliding window_.
**Example 1:**
**Input:** nums... | How about using a data structure such as deque (double-ended queue)? The queue size need not be the same as the window’s size. Remove redundant elements and the queue should store only elements that need to be considered. |
Python || Sliding window || beats 99.81% | sliding-window-maximum | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWhenever you see something related to subarrays, sliding window is something that always comes to mind. So here too as we have to find maximum in each window of size k, we have to \n- Maintain a window. \n- Find maximum element in that wi... | 0 | You are given an array of integers `nums`, there is a sliding window of size `k` which is moving from the very left of the array to the very right. You can only see the `k` numbers in the window. Each time the sliding window moves right by one position.
Return _the max sliding window_.
**Example 1:**
**Input:** nums... | How about using a data structure such as deque (double-ended queue)? The queue size need not be the same as the window’s size. Remove redundant elements and the queue should store only elements that need to be considered. |
Simple Heap(Priority Queue) Approach | sliding-window-maximum | 0 | 1 | \n\n# Complexity\n- Time complexity:\n$$O(nlogn)$$\n\n- Space complexity:\n$$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nimport heapq\n\nclass Solution:\n def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:\n if k == 1:\n return nums\n\n st = ... | 1 | You are given an array of integers `nums`, there is a sliding window of size `k` which is moving from the very left of the array to the very right. You can only see the `k` numbers in the window. Each time the sliding window moves right by one position.
Return _the max sliding window_.
**Example 1:**
**Input:** nums... | How about using a data structure such as deque (double-ended queue)? The queue size need not be the same as the window’s size. Remove redundant elements and the queue should store only elements that need to be considered. |
Simple Python Solution - Easy to understand | sliding-window-maximum | 0 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\n1. Loop through all the elements\n2. Keep on pushing (-number, index) into the heap (- for max heap)\n3. Pop of all the max elements from the heap whose index is less than i-k (we should only consider i-k to i elements everytime)\n4. If index is more ... | 2 | You are given an array of integers `nums`, there is a sliding window of size `k` which is moving from the very left of the array to the very right. You can only see the `k` numbers in the window. Each time the sliding window moves right by one position.
Return _the max sliding window_.
**Example 1:**
**Input:** nums... | How about using a data structure such as deque (double-ended queue)? The queue size need not be the same as the window’s size. Remove redundant elements and the queue should store only elements that need to be considered. |
Beats 100% | Easy To Understand | 8 Lines of Code 🚀🚀 | sliding-window-maximum | 1 | 1 | # Intuition\n\n\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 com... | 2 | You are given an array of integers `nums`, there is a sliding window of size `k` which is moving from the very left of the array to the very right. You can only see the `k` numbers in the window. Each time the sliding window moves right by one position.
Return _the max sliding window_.
**Example 1:**
**Input:** nums... | How about using a data structure such as deque (double-ended queue)? The queue size need not be the same as the window’s size. Remove redundant elements and the queue should store only elements that need to be considered. |
🔥 Fast O(n) Deque | sliding-window-maximum | 0 | 1 | # Intuition\nAt first glance, the problem seems to be a simple maximum calculation problem within a range. However, as the range, or the window, slides through the array, recalculating the maximum for each new window seems inefficient. Our goal is to find an approach where we can leverage the previous window\'s informa... | 8 | You are given an array of integers `nums`, there is a sliding window of size `k` which is moving from the very left of the array to the very right. You can only see the `k` numbers in the window. Each time the sliding window moves right by one position.
Return _the max sliding window_.
**Example 1:**
**Input:** nums... | How about using a data structure such as deque (double-ended queue)? The queue size need not be the same as the window’s size. Remove redundant elements and the queue should store only elements that need to be considered. |
[Python3] Decreasing queue || beats 95% || 1296ms 🥷🏼 | sliding-window-maximum | 0 | 1 | ```python3 []\nclass Solution:\n def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:\n q, res = deque(), [] # save index in the queue \'q\' (decreasing order)\n for r in range(len(nums)):\n # remove from the right side of the queue all items less than current\n while... | 11 | You are given an array of integers `nums`, there is a sliding window of size `k` which is moving from the very left of the array to the very right. You can only see the `k` numbers in the window. Each time the sliding window moves right by one position.
Return _the max sliding window_.
**Example 1:**
**Input:** nums... | How about using a data structure such as deque (double-ended queue)? The queue size need not be the same as the window’s size. Remove redundant elements and the queue should store only elements that need to be considered. |
Simplest O(n) Python Solution with Explanation | sliding-window-maximum | 0 | 1 | The solution using deque isn\'t very intuitive but is very simple once it *"clicks"*\n\n\nTo summarize:\n1. We maintain a deque of the indexes of the largest elements we\'ve seen (aka "good candidates")\n2. The problem is that we need to maintain sanity of this deque. To this we need to make sure about two things:\n\t1... | 177 | You are given an array of integers `nums`, there is a sliding window of size `k` which is moving from the very left of the array to the very right. You can only see the `k` numbers in the window. Each time the sliding window moves right by one position.
Return _the max sliding window_.
**Example 1:**
**Input:** nums... | How about using a data structure such as deque (double-ended queue)? The queue size need not be the same as the window’s size. Remove redundant elements and the queue should store only elements that need to be considered. |
:) | search-a-2d-matrix-ii | 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 | Write an efficient algorithm that searches for a value `target` in an `m x n` integer matrix `matrix`. This matrix has the following properties:
* Integers in each row are sorted in ascending from left to right.
* Integers in each column are sorted in ascending from top to bottom.
**Example 1:**
**Input:** matri... | null |
✔️ PYTHON || EXPLAINED || ; ] | search-a-2d-matrix-ii | 0 | 1 | **UPVOTE IF HELPFuuL**\n\n**EFFICIENT APPROACH -> O ( M + N )**\n\nAs the rows are sorted **->** ```matrix[i][j] < matrix[i][j+1]```\nAs the columns are sorted **->** ```matrix[i][j] >matrix[i-1][j]```\n\nHence it can be said that :\n* any element right to ```matrix[i][j]``` will be greater than it.\n* any element to t... | 83 | Write an efficient algorithm that searches for a value `target` in an `m x n` integer matrix `matrix`. This matrix has the following properties:
* Integers in each row are sorted in ascending from left to right.
* Integers in each column are sorted in ascending from top to bottom.
**Example 1:**
**Input:** matri... | null |
✅Easy Solution with Explanation | 2 Approaches 😀👍🏻🚀 | search-a-2d-matrix-ii | 0 | 1 | # Video Explanation\nhttps://youtu.be/QMs447Imtdc\n\n#### Upvote if find useful \uD83D\uDE00\uD83D\uDC4D\uD83C\uDFFB\uD83D\uDE80\n\n# Approach 1\nUsing Binary Search\n\n# Complexity\n- Time complexity:\nO(mlogn)\n\n# Code\n```\nclass Solution:\n def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:\n... | 2 | Write an efficient algorithm that searches for a value `target` in an `m x n` integer matrix `matrix`. This matrix has the following properties:
* Integers in each row are sorted in ascending from left to right.
* Integers in each column are sorted in ascending from top to bottom.
**Example 1:**
**Input:** matri... | null |
Awesome Logic With Time Complexity---->O(N) | search-a-2d-matrix-ii | 0 | 1 | \n\n# Normal Approach---->O(N)\n```\nclass Solution:\n def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:\n row,col=len(matrix),len(matrix[0])\n r,c=0,col-1\n while r<row and c>=0:\n if matrix[r][c]==target:\n return True\n if matrix[r][c]>... | 14 | Write an efficient algorithm that searches for a value `target` in an `m x n` integer matrix `matrix`. This matrix has the following properties:
* Integers in each row are sorted in ascending from left to right.
* Integers in each column are sorted in ascending from top to bottom.
**Example 1:**
**Input:** matri... | null |
Python3 Brute force approach Beats 69.25% (Time) and Beats 86.20%(Space) | search-a-2d-matrix-ii | 0 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\nIterate over the rows of the matrix and apply binary search on each of the row matrices to search the element\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n O(n*log(m))\n\n- Space complexity:\n<!-- Add yo... | 1 | Write an efficient algorithm that searches for a value `target` in an `m x n` integer matrix `matrix`. This matrix has the following properties:
* Integers in each row are sorted in ascending from left to right.
* Integers in each column are sorted in ascending from top to bottom.
**Example 1:**
**Input:** matri... | null |
Python Easy Solution || 100% || Beats 96% || Binary Search || | search-a-2d-matrix-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nEach Row is sorted, We will use Binary Search on Each Row\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O(n(log(n)))$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n... | 2 | Write an efficient algorithm that searches for a value `target` in an `m x n` integer matrix `matrix`. This matrix has the following properties:
* Integers in each row are sorted in ascending from left to right.
* Integers in each column are sorted in ascending from top to bottom.
**Example 1:**
**Input:** matri... | null |
Most optimal solution with explanation | Eliminating the rows and cols (concept of binary search) | search-a-2d-matrix-ii | 1 | 1 | \n\n# Approach\n1. Start from the top-right corner of the matrix (row = 0, col = m-1), where n is the number of rows and m is the number of columns in the matrix.\n2. Compare the element at the current position (matrix[row][col]) with the target value:\n - If the element is equal to the target, return true.\n - I... | 7 | Write an efficient algorithm that searches for a value `target` in an `m x n` integer matrix `matrix`. This matrix has the following properties:
* Integers in each row are sorted in ascending from left to right.
* Integers in each column are sorted in ascending from top to bottom.
**Example 1:**
**Input:** matri... | null |
Searching 2d Simple Solution | search-a-2d-matrix-ii | 0 | 1 | \n\n# Superb Logical Solution in Python\n```\nclass Solution:\n def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:\n for row in matrix:\n if target in row:\n return True\n\n return False\n```\n# please upvote me it would encourages me | 2 | Write an efficient algorithm that searches for a value `target` in an `m x n` integer matrix `matrix`. This matrix has the following properties:
* Integers in each row are sorted in ascending from left to right.
* Integers in each column are sorted in ascending from top to bottom.
**Example 1:**
**Input:** matri... | null |
240: Time 91.4% and Space 98.52%, Solution with step by step explanation | search-a-2d-matrix-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThe problem can be solved using a divide and conquer approach. We start from the top right corner and move towards the bottom left corner of the matrix. At each position, we check if the current element is equal to the targe... | 13 | Write an efficient algorithm that searches for a value `target` in an `m x n` integer matrix `matrix`. This matrix has the following properties:
* Integers in each row are sorted in ascending from left to right.
* Integers in each column are sorted in ascending from top to bottom.
**Example 1:**
**Input:** matri... | null |
Python3 || 9 lines, w/ explanation || T/M: 84%/82% | search-a-2d-matrix-ii | 0 | 1 | ```\nclass Solution: # Here\'s the plan:\n # \u2022 We start in the bottom-left corner, which is \n # matrix[-1][0]. Obviously, the target cell cannot\n # be either below or to the left of this corner cell\n # ... | 18 | Write an efficient algorithm that searches for a value `target` in an `m x n` integer matrix `matrix`. This matrix has the following properties:
* Integers in each row are sorted in ascending from left to right.
* Integers in each column are sorted in ascending from top to bottom.
**Example 1:**
**Input:** matri... | null |
Recursion -> Memoization -> Tabulation | Partition DP | C++, Java, Python | With Explanation | different-ways-to-add-parentheses | 1 | 1 | # Intuition\nWhen we try to evaluate a operator, we again encounter the same problem on the left & right substring of the operator & as we have to calculate all possible ways, that\'s where recursion comes in picture.\n\n# Recursion\nWe define a recursive function `getDiffWays` where, `getDiffWays(i, j)` returns us num... | 14 | Given a string `expression` of numbers and operators, return _all possible results from computing all the different possible ways to group numbers and operators_. You may return the answer in **any order**.
The test cases are generated such that the output values fit in a 32-bit integer and the number of different res... | null |
Recursive python solution using eval | different-ways-to-add-parentheses | 0 | 1 | # Code\n```\nclass Solution:\n def diffWaysToCompute(self, expression: str) -> List[int]:\n result = []\n operators = "*-+"\n isNumber = True\n for op in operators:\n if op in expression:\n isNumber = False\n if isNumber:\n return [int(expressio... | 4 | Given a string `expression` of numbers and operators, return _all possible results from computing all the different possible ways to group numbers and operators_. You may return the answer in **any order**.
The test cases are generated such that the output values fit in a 32-bit integer and the number of different res... | null |
Easy understand D&C solution, no helper function needed, Pattern example | different-ways-to-add-parentheses | 0 | 1 | The most difficult part is finding the pattern\nI will set example to show the pattern I found:\nLet\'s say 2-1-1+1, you can think it like a tree, every time when you read the operator you will split it into left part and right part:\n\n```\niteration 1:\n\t\t\t2 - 1- 1 + 1\n\t\t / \\\n\t\t 2 - (1-1+1)\n\t\t /\t ... | 65 | Given a string `expression` of numbers and operators, return _all possible results from computing all the different possible ways to group numbers and operators_. You may return the answer in **any order**.
The test cases are generated such that the output values fit in a 32-bit integer and the number of different res... | null |
241: Solution with step by step explanation | different-ways-to-add-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nWe can use divide and conquer approach to solve this problem. We can divide the given expression into two parts at each operator, and recursively calculate the results for each part. Finally, we can combine the results using... | 8 | Given a string `expression` of numbers and operators, return _all possible results from computing all the different possible ways to group numbers and operators_. You may return the answer in **any order**.
The test cases are generated such that the output values fit in a 32-bit integer and the number of different res... | null |
Python: Divde-and-Conquer + Memoization + O(N * 2^N) | different-ways-to-add-parentheses | 0 | 1 | Hello,\n\nThe time complexity analysis is below.\n\n```python\n\ndef __init__(self):\n\tself.operation = {}\n\tself.operation[\'*\'] = lambda a, b: a * b\n\tself.operation[\'+\'] = lambda a, b: a + b\n\tself.operation[\'-\'] = lambda a, b: a - b\n\ndef diffWaysToCompute(self, expression: str) -> List[int]:\n\n\tmemo = ... | 34 | Given a string `expression` of numbers and operators, return _all possible results from computing all the different possible ways to group numbers and operators_. You may return the answer in **any order**.
The test cases are generated such that the output values fit in a 32-bit integer and the number of different res... | null |
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