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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Python3 Solution | candy | 0 | 1 | \n```\nclass Solution:\n def candy(self, ratings: List[int]) -> int:\n n=len(ratings)\n dp=[1]*n\n for i in range(1,n):\n if ratings[i]>ratings[i-1]:\n dp[i]=dp[i-1]+1\n\n for i in range(n-2,-1,-1):\n if ratings[i]>ratings[i+1]:\n dp[i]=... | 2 | There are `n` children standing in a line. Each child is assigned a rating value given in the integer array `ratings`.
You are giving candies to these children subjected to the following requirements:
* Each child must have at least one candy.
* Children with a higher rating get more candies than their neighbors.... | null |
Python 3 | Beats 88.46% Time and 99.36% Memory | 2 Pass solution | candy | 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 | There are `n` children standing in a line. Each child is assigned a rating value given in the integer array `ratings`.
You are giving candies to these children subjected to the following requirements:
* Each child must have at least one candy.
* Children with a higher rating get more candies than their neighbors.... | null |
🚀 Beats 99.93% || Greedy || Two Solutions || C++ || Java || Python || Commented Code 🚀 | candy | 1 | 1 | # Problem Description\nThe task is distributing candies to a group of `n` children standing in a line. Each child is assigned a **rating value**, represented as an **integer** array called `ratings`.\n\nFollow **two specific requirements** while distributing the candies:\n- Each child must have at least one candy.\n- C... | 76 | There are `n` children standing in a line. Each child is assigned a rating value given in the integer array `ratings`.
You are giving candies to these children subjected to the following requirements:
* Each child must have at least one candy.
* Children with a higher rating get more candies than their neighbors.... | null |
【Video】O(n) Time, O(1) Space Solution - Python, JavaScript, Java, C++ | candy | 1 | 1 | # Intuition\nKeep two peak values and subtract the lower peak from total.\n\n---\n\n# Solution Video\n\nhttps://youtu.be/f5oFx-X0eS4\n\n### \u2B50\uFE0F\u2B50\uFE0F Don\'t forget to subscribe to my channel! \u2B50\uFE0F\u2B50\uFE0F\n\n**\u25A0 Subscribe URL**\nhttp://www.youtube.com/channel/UC9RMNwYTL3SXCP6ShLWVFww?sub... | 32 | There are `n` children standing in a line. Each child is assigned a rating value given in the integer array `ratings`.
You are giving candies to these children subjected to the following requirements:
* Each child must have at least one candy.
* Children with a higher rating get more candies than their neighbors.... | null |
Simple, extremely readable O(n) time O(1) space | candy | 0 | 1 | # Intuition\nSee Editorial for intuition regarding slope approach (Approach 4: Single Pass Approach with Constant Space). \nSee section below code for notes on my approach.\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution:\n def candy(self, ratings: List[int]) -> ... | 4 | There are `n` children standing in a line. Each child is assigned a rating value given in the integer array `ratings`.
You are giving candies to these children subjected to the following requirements:
* Each child must have at least one candy.
* Children with a higher rating get more candies than their neighbors.... | null |
Beginner Friendly || Line By Line Explanation || Easy Candy Solution || Python || Java || Beats 95%+ | candy | 1 | 1 | # Beats \n\n\n# JAVA CODE 1-2ms \n# UPVOTE IF U LIKE !!!\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe main intuition behind this problem is to ensure that children... | 27 | There are `n` children standing in a line. Each child is assigned a rating value given in the integer array `ratings`.
You are giving candies to these children subjected to the following requirements:
* Each child must have at least one candy.
* Children with a higher rating get more candies than their neighbors.... | null |
Python | 99.82% | Beginner Friendly | Optimal Solution | candy | 0 | 1 | # Python | 99.82% | Beginner Friendly | Optimal Solution\n```\nclass Solution:\n def candy(self, R):\n n, ans = len(R), [1]*len(R)\n \n for i in range(n-1):\n if R[i] < R[i+1]:\n ans[i+1] = max(1 + ans[i], ans[i+1])\n \n for i in range(n-2, -1, -1)... | 19 | There are `n` children standing in a line. Each child is assigned a rating value given in the integer array `ratings`.
You are giving candies to these children subjected to the following requirements:
* Each child must have at least one candy.
* Children with a higher rating get more candies than their neighbors.... | null |
Python simple two-pass: forth and back | candy | 0 | 1 | # Intuition\nSome other solutions are too much complicated, which shouldn\'t be. Just play some cases to get this idea. Please up vote if you agree with this idea and let me see if you like it.\n\nOkay, just saw other friends already got this idea. I did work it out alone since it\'s pretty straightforward. Anyways, I ... | 24 | There are `n` children standing in a line. Each child is assigned a rating value given in the integer array `ratings`.
You are giving candies to these children subjected to the following requirements:
* Each child must have at least one candy.
* Children with a higher rating get more candies than their neighbors.... | null |
Python Easy Code! Beats 95.82% Runtime and 94.23% Memory! | candy | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nI used the approach of once iterating from left to right and then from right to left. Here, we had to consider kids on either sides, hence this was a good technique to achieve O(n).\n# Approach\n<!-- Describe your approach to solving the ... | 6 | There are `n` children standing in a line. Each child is assigned a rating value given in the integer array `ratings`.
You are giving candies to these children subjected to the following requirements:
* Each child must have at least one candy.
* Children with a higher rating get more candies than their neighbors.... | null |
Greedy || C++ || Java || Python || Commented Code | candy | 1 | 1 | # Intuition && approach\n<!-- -->\nThe first step is to create a vector of integers called candies with the same size as the ratings array. Each element of candies will store the number of candies that the corresponding child should get. We initialize all elements of candies to 1.\n\nIn the next step, we iterate throu... | 5 | There are `n` children standing in a line. Each child is assigned a rating value given in the integer array `ratings`.
You are giving candies to these children subjected to the following requirements:
* Each child must have at least one candy.
* Children with a higher rating get more candies than their neighbors.... | null |
✅ EASY | ✅ [ Python / Java / C++ / JavaScript / C# ] | 🔥100 % | | candy | 1 | 1 | \n```Python []\nclass Solution:\n def candy(self, ratings: List[int]) -> int:\n n = len(ratings)\n if n <= 1:\n return n\n\n nums = [1] * n\n\n # Make sure children with a higher rating get more candy than their left neighbor\n for i in range(1, n):\n if ratin... | 4 | There are `n` children standing in a line. Each child is assigned a rating value given in the integer array `ratings`.
You are giving candies to these children subjected to the following requirements:
* Each child must have at least one candy.
* Children with a higher rating get more candies than their neighbors.... | null |
Detailed Explanation | candy | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nGiven, the person with highest rating compared to neighbours should have more candies.There are obviously two possiblities\n1.The rating is more than the rating of the left person (if present)the candies will be one more than the candies ... | 1 | There are `n` children standing in a line. Each child is assigned a rating value given in the integer array `ratings`.
You are giving candies to these children subjected to the following requirements:
* Each child must have at least one candy.
* Children with a higher rating get more candies than their neighbors.... | null |
Think it through || Time: O(n) Space: O(1) || Python Go Explained | single-number | 0 | 1 | \n### Edge Cases:\n1. No element appears twice; it is a constraint so not possible\n2. Single length array; return the only element already present in the array\n3. len(nums) > 1; find the single element that does not appear twice\n\n### Approaches:\n1. **Brute Force**\nIntuition:\nIterate through every element in the ... | 723 | Given a **non-empty** array of integers `nums`, every element appears _twice_ except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,1\]
**Output:** 1
**Example 2:**
**Input:** nums = \[4,1,2,1,2... | null |
✔️ [Python3] ONE-LINER ヾ（*⌒ヮ⌒*）ゞ, Explained | single-number | 0 | 1 | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nWe use the nice property of XOR operation which is if you XOR same numbers it will return zero. Since the `nums` contains just one non-repeating number, we can just XOR all numbers together and the final result will ... | 193 | Given a **non-empty** array of integers `nums`, every element appears _twice_ except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,1\]
**Output:** 1
**Example 2:**
**Input:** nums = \[4,1,2,1,2... | null |
Solution | single-number | 1 | 1 | ```C++ []\nclass Solution {\npublic:\n int singleNumber(vector<int>& nums) {\n \n int ans = nums[0];\n\n for(int i = 1 ; i < nums.size() ; i++){\n ans = ans ^ nums[i];\n }\n\n return ans;\n }\n};\n```\n\n```Python3 []\nclass Solution:\n def singleNumber(self, nums: L... | 474 | Given a **non-empty** array of integers `nums`, every element appears _twice_ except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,1\]
**Output:** 1
**Example 2:**
**Input:** nums = \[4,1,2,1,2... | null |
beats everyone | single-number | 0 | 1 | \n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def singleNumber(self, nums: List[int]) -> int:\n dic = {}\n\n for i in range(len(nums)):\n if nums[i] not in dic:\n dic[nums[i]] = 1\n else:\n dic.pop(nums[i])... | 1 | Given a **non-empty** array of integers `nums`, every element appears _twice_ except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,1\]
**Output:** 1
**Example 2:**
**Input:** nums = \[4,1,2,1,2... | null |
Python code with Time Complexity: T(O(N)) and space Complexity: T(O(M)) | single-number | 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 a **non-empty** array of integers `nums`, every element appears _twice_ except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,1\]
**Output:** 1
**Example 2:**
**Input:** nums = \[4,1,2,1,2... | null |
Python3 | Using Dictionary | single-number | 0 | 1 | \n# Approach\n- Create dict\n- Iterate through \'nums\' list\n- If an element is not in our dict -> add new key in dict\n- Else (Element is duplicated) -> `pop` it from dict\n- Unpack dict using `*` and return it\n\n# Complexity\n- Time complexity: The time complexity of the provided code is O(n)\n Beats 60.16% of user... | 1 | Given a **non-empty** array of integers `nums`, every element appears _twice_ except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,1\]
**Output:** 1
**Example 2:**
**Input:** nums = \[4,1,2,1,2... | null |
Most Fastest Python Solution | single-number | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSolve the Problem easily and fastly\n\n\n\n# Approach\n<!-- Describe your approach to solving the pro... | 2 | Given a **non-empty** array of integers `nums`, every element appears _twice_ except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,1\]
**Output:** 1
**Example 2:**
**Input:** nums = \[4,1,2,1,2... | null |
Very Easy || 0 ms ||100%|| Fully Explained (Java, C++, Python, JS, C, Python3) | single-number | 1 | 1 | **Problem:**\nGiven a non-empty array of integers nums, every element appears twice except for one. Find that single one.\n**Input:** nums = [ 4, 1, 2, 1, 2 ]\n**Output:** 4\n**Explanation:** 1\u2019s and 2\u2019s appear twice, only 4 appears exactly once. So the answer is 4.\n**Concept of XOR:**\nXOR of zero and ... | 150 | Given a **non-empty** array of integers `nums`, every element appears _twice_ except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,1\]
**Output:** 1
**Example 2:**
**Input:** nums = \[4,1,2,1,2... | null |
Single-number | single-number | 0 | 1 | # Intuition\n<!-- The code aims to find and return the single number that appears only once in a list of numbers.\n -->\n\n# Approach\n<!-- 1.Utilize a dictionary (mydict) to count the occurrences of each number in the given list.\n2. Iterate through the dictionary and return the number with an occurrence of 1. -->\n\n... | 1 | Given a **non-empty** array of integers `nums`, every element appears _twice_ except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,1\]
**Output:** 1
**Example 2:**
**Input:** nums = \[4,1,2,1,2... | null |
Single Number-136(Solution) | single-number | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTry to solve first itself.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. nums list converted into set.\n2. Find the ocurrance of each element using the count() function.\n\n# Complexity\n- Time complexity:\n<!... | 2 | Given a **non-empty** array of integers `nums`, every element appears _twice_ except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,1\]
**Output:** 1
**Example 2:**
**Input:** nums = \[4,1,2,1,2... | null |
Easy XOR soultion | single-number | 1 | 1 | # Intuition\nI solved it via XOR bit operator\n\n# Approach\nFor instance, I have an array [12, 1, 13, 1, 12]\n1010\n0001\n\n1011\n1011\n\n0000\n0001\n\n0001\n1010\n\n1011 (bin) = 13 (dec)\n\n# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# C... | 11 | Given a **non-empty** array of integers `nums`, every element appears _twice_ except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,1\]
**Output:** 1
**Example 2:**
**Input:** nums = \[4,1,2,1,2... | null |
Python shortest 1-liner. Functional programming. | single-number | 0 | 1 | # Approach\n1. Xor of a number with itself gives zero, i.e `a ^ a = 0`\n\n2. Xor of a number with zero gives back the number as is, i.e `a ^ 0 = a`\n\n3. Using the above two properties, xor all the numbers in `nums`.\n The numbers which appear twice will be zeroed, leaving the single occuring number at the end.\n\n#... | 4 | Given a **non-empty** array of integers `nums`, every element appears _twice_ except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,1\]
**Output:** 1
**Example 2:**
**Input:** nums = \[4,1,2,1,2... | null |
Beats 100% | Easy To Understand | 0ms 🔥🚀 | single-number | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nXor of any number with itself is 1, so the duplicate numbers will become 1 and we will get the unique number.\n\n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complex... | 3 | Given a **non-empty** array of integers `nums`, every element appears _twice_ except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,1\]
**Output:** 1
**Example 2:**
**Input:** nums = \[4,1,2,1,2... | null |
Python 3 diff approach easy to understand 3 line code | single-number | 0 | 1 | **Plz Upvote ..if you got help from this.**\n# Code\n```\nclass Solution:\n def singleNumber(self, nums: List[int]) -> int:\n<!-- SIMPLE APPROACH 90 %BEATS -->\n<!-- ===================================================== -->\n for i in range(len(nums)):\n if(nums.count(nums[i])==1):\n ... | 8 | Given a **non-empty** array of integers `nums`, every element appears _twice_ except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,1\]
**Output:** 1
**Example 2:**
**Input:** nums = \[4,1,2,1,2... | null |
✅Bit Manipulation🔥 || C++ || JAVA || PYTHON || Beginner Friendly🔥🔥🔥 | single-number-ii | 1 | 1 | # Approach 1: Brute Force\n\n# Code\n```C++ []\nclass Solution {\npublic:\n int singleNumber(vector<int>& nums) {\n unordered_map<int, int> m;\n \n for(auto x: nums){\n m[x]++;\n }\n\n for(auto x: m){\n if(x.second == 1){\n return x.first;\n ... | 538 | Given an integer array `nums` where every element appears **three times** except for one, which appears **exactly once**. _Find the single element and return it_.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,3,2\]
**Output:**... | null |
Time complexity O(n) and space complexity O(n) | single-number-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- O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n- O(n)\n<!-- Add your space complexity here, ... | 1 | Given an integer array `nums` where every element appears **three times** except for one, which appears **exactly once**. _Find the single element and return it_.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,3,2\]
**Output:**... | null |
Python Simple Counter Solution | Linear runtime | single-number-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:\nO(n)\n\n- Space complexity:\nO(n / 3)\n\n# Code\n```\nfrom collections import Counter\n\nclass Solution:\n def singleNumber(self... | 1 | Given an integer array `nums` where every element appears **three times** except for one, which appears **exactly once**. _Find the single element and return it_.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,3,2\]
**Output:**... | null |
Python very easy Solution || Use Hashmap | single-number-ii | 0 | 1 | # Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n\n# Code\n```\nclass Solution:\n def singleNumber(self, nums: List[int]) -> int:\n dic=dict()\n for i in nums:\n if i in dic:\n dic[i]+=1\n if dic[i]==3:\n dic.pop(i)\n ... | 1 | Given an integer array `nums` where every element appears **three times** except for one, which appears **exactly once**. _Find the single element and return it_.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,3,2\]
**Output:**... | null |
Image Explanation🏆- [Bit Manipulation - 4 Methods] - C++/Java/Python | single-number-ii | 1 | 1 | # Video Solution (`Aryan Mittal`) - Link in LeetCode Profile\n`Single Number II` by `Aryan Mittal`\n\n\n\n# Approach & Intution\n$$\n\n- Space complexity: $$O(n)$$\n\nwhere, `n is length of nums`.\n\n# Code\n```python\nclass Solution:\n def singleNumber(self, nums: list[int]) -> int:\n def counts(a: tuple[int, int], nu... | 1 | Given an integer array `nums` where every element appears **three times** except for one, which appears **exactly once**. _Find the single element and return it_.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,3,2\]
**Output:**... | null |
Single line mathematical solution, beats 100% | single-number-ii | 0 | 1 | # Approach\n- Find all the distinct numbers (set), multiply by 3 (because all the other number repeats 3 time), so the sum of that set will have total of original nums + 2*(number which is repeating only once)\n- Now substract the sum of original list of nums, which gives total as 2*(number which repeats only once)\n\n... | 16 | Given an integer array `nums` where every element appears **three times** except for one, which appears **exactly once**. _Find the single element and return it_.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,3,2\]
**Output:**... | null |
Bitwise Operations | Python / JS Solution | single-number-ii | 0 | 1 | Hello **Tenno leetcoders**, \n\nWe are given an integer array nums where every element appears three times except for one, which appears exactly once. Find the single element and return a solution with a linear runtime complexity and use only constant extra space.\n\n### Explanation\n\nThe problem requires finding the ... | 9 | Given an integer array `nums` where every element appears **three times** except for one, which appears **exactly once**. _Find the single element and return it_.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,3,2\]
**Output:**... | null |
PYTHON EASY SOLUTION || 100% | single-number-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. $$O(n)$$ --... | 1 | Given an integer array `nums` where every element appears **three times** except for one, which appears **exactly once**. _Find the single element and return it_.
You must implement a solution with a linear runtime complexity and use only constant extra space.
**Example 1:**
**Input:** nums = \[2,2,3,2\]
**Output:**... | null |
✅ 97.92% Hash Table & Linked List | copy-list-with-random-pointer | 1 | 1 | # Interview Guide - Copying a Linked List with Random Pointers: A Dual-Approach Analysis\n\n## Introduction & Problem Understanding\n\nThe problem at hand involves creating a deep copy of a given singly-linked list where each node has a `next` pointer and an additional `random` pointer. The `random` pointer could point... | 188 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
✅ 97.92% Hash Table & Linked List | copy-list-with-random-pointer | 1 | 1 | # Interview Guide - Copying a Linked List with Random Pointers: A Dual-Approach Analysis\n\n## Introduction & Problem Understanding\n\nThe problem at hand involves creating a deep copy of a given singly-linked list where each node has a `next` pointer and an additional `random` pointer. The `random` pointer could point... | 188 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Beginner-friendly || Solution with using Hashtable + LinkedList DS in Python3 | copy-list-with-random-pointer | 0 | 1 | # Intuition\nThe problem description is the following:\n- we have a **linked list** of nodes\n- we shall perform **a deep copy** == this means, that **all of the values and pointers** MUST BE **copied with a new referrence/value for storing** (primitives are passing by **value**, and objects - **by referrence**)\n\nLet... | 1 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Beginner-friendly || Solution with using Hashtable + LinkedList DS in Python3 | copy-list-with-random-pointer | 0 | 1 | # Intuition\nThe problem description is the following:\n- we have a **linked list** of nodes\n- we shall perform **a deep copy** == this means, that **all of the values and pointers** MUST BE **copied with a new referrence/value for storing** (primitives are passing by **value**, and objects - **by referrence**)\n\nLet... | 1 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Optimized and Easy-to-Understand Copy Random Linked List - Java, C++, Python | copy-list-with-random-pointer | 1 | 1 | ## Intuition\nWhen faced with the problem of copying a linked list with random pointers, the initial thought is to use a hashmap to keep track of the mapping between the original nodes and their corresponding copies. This way, we can efficiently clone the linked list while handling the random pointers correctly.\n\n## ... | 1 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Optimized and Easy-to-Understand Copy Random Linked List - Java, C++, Python | copy-list-with-random-pointer | 1 | 1 | ## Intuition\nWhen faced with the problem of copying a linked list with random pointers, the initial thought is to use a hashmap to keep track of the mapping between the original nodes and their corresponding copies. This way, we can efficiently clone the linked list while handling the random pointers correctly.\n\n## ... | 1 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
⛳ Python3 🧲 📣 Beats 🔥 99.25% 📣 🧲 | copy-list-with-random-pointer | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n\n# Approach\n1. First Create the list with only next pointer\n2. Store Address of the actual list corresponds to copy lis... | 1 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
⛳ Python3 🧲 📣 Beats 🔥 99.25% 📣 🧲 | copy-list-with-random-pointer | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n\n# Approach\n1. First Create the list with only next pointer\n2. Store Address of the actual list corresponds to copy lis... | 1 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Hardest Python3 solution. | copy-list-with-random-pointer | 0 | 1 | # Code: \n\n```\nclass Solution:\n def copyRandomList(self, head):\n return __import__("copy").deepcopy(head)\n``` | 1 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Hardest Python3 solution. | copy-list-with-random-pointer | 0 | 1 | # Code: \n\n```\nclass Solution:\n def copyRandomList(self, head):\n return __import__("copy").deepcopy(head)\n``` | 1 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
✅Clear Explanation🔥&Code✅||🔥Step-by-Step Guide to Deep Copying🔥 | copy-list-with-random-pointer | 1 | 1 | # Problem Description:\n\nYou are given a **linked list** where each node has a **"random"** pointer that can **point to any other node in the list**, or it can be null. You need to **create a deep copy of this linked list**, such that the **copied list has new nodes with the same values as the original list**, and the... | 41 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
✅Clear Explanation🔥&Code✅||🔥Step-by-Step Guide to Deep Copying🔥 | copy-list-with-random-pointer | 1 | 1 | # Problem Description:\n\nYou are given a **linked list** where each node has a **"random"** pointer that can **point to any other node in the list**, or it can be null. You need to **create a deep copy of this linked list**, such that the **copied list has new nodes with the same values as the original list**, and the... | 41 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
【Video】Solution with HashMap - Python, JavaScript, Java, C++ | copy-list-with-random-pointer | 1 | 1 | # Intuition\nThe main idea to solve the question of copying a linked list with random pointers is to create a deep copy of the original linked list while maintaining the relationships between nodes. By following these steps below, you can create an exact deep copy of the original linked list, including its random point... | 25 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
【Video】Solution with HashMap - Python, JavaScript, Java, C++ | copy-list-with-random-pointer | 1 | 1 | # Intuition\nThe main idea to solve the question of copying a linked list with random pointers is to create a deep copy of the original linked list while maintaining the relationships between nodes. By following these steps below, you can create an exact deep copy of the original linked list, including its random point... | 25 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Python short and clean. Single pass. | copy-list-with-random-pointer | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n\n- Space complexity: $$O(n)$$\n\nwhere, `n is number of nodes in LinkedList starting at head`.\n\n# Code\n```python\nclass Solution:\n def copyRandomList(self, head: \'Optional[Node]\') -> \'Optional[Node]\':\n copy = defaultdict(lambda: Node(0), {None: None})\n ... | 1 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Python short and clean. Single pass. | copy-list-with-random-pointer | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n\n- Space complexity: $$O(n)$$\n\nwhere, `n is number of nodes in LinkedList starting at head`.\n\n# Code\n```python\nclass Solution:\n def copyRandomList(self, head: \'Optional[Node]\') -> \'Optional[Node]\':\n copy = defaultdict(lambda: Node(0), {None: None})\n ... | 1 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Python3 | Easy to Understand | Mapping | copy-list-with-random-pointer | 0 | 1 | # Python3 | Easy to Understand | Mapping\n```\n"""\n# Definition for a Node.\nclass Node:\n def __init__(self, val, next, random):\n self.val = val\n self.next = next\n self.random = random\n"""\nclass Solution:\n def copyRandomList(self, head: \'Node\') -> \'Node\':\n if head is None:... | 2 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Python3 | Easy to Understand | Mapping | copy-list-with-random-pointer | 0 | 1 | # Python3 | Easy to Understand | Mapping\n```\n"""\n# Definition for a Node.\nclass Node:\n def __init__(self, val, next, random):\n self.val = val\n self.next = next\n self.random = random\n"""\nclass Solution:\n def copyRandomList(self, head: \'Node\') -> \'Node\':\n if head is None:... | 2 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Python | Easy Solution | copy-list-with-random-pointer | 0 | 1 | # Intuition\nUsing hashed Linked List\n\n# Approach\nTo create a deep copy of a linked list with random pointers, you can follow these steps:\n\n1. Iterate through the original linked list, creating a new node for each node in the original list and mapping the original node to its corresponding new node in a dictionary... | 1 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Python | Easy Solution | copy-list-with-random-pointer | 0 | 1 | # Intuition\nUsing hashed Linked List\n\n# Approach\nTo create a deep copy of a linked list with random pointers, you can follow these steps:\n\n1. Iterate through the original linked list, creating a new node for each node in the original list and mapping the original node to its corresponding new node in a dictionary... | 1 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Beats 99%✅ | O( n )✅ | Easy-To-Understand✅ | Python (Step by step explanation)✅ | copy-list-with-random-pointer | 0 | 1 | # Intuition\nThe intuition for solving this problem is to create a deep copy of the linked list with random pointers. We can achieve this by using a hashmap to map the original nodes to their corresponding new nodes.\n\n# Approach\n1. Create a hashmap `mapp` to map the original nodes to their corresponding new nodes. I... | 2 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Beats 99%✅ | O( n )✅ | Easy-To-Understand✅ | Python (Step by step explanation)✅ | copy-list-with-random-pointer | 0 | 1 | # Intuition\nThe intuition for solving this problem is to create a deep copy of the linked list with random pointers. We can achieve this by using a hashmap to map the original nodes to their corresponding new nodes.\n\n# Approach\n1. Create a hashmap `mapp` to map the original nodes to their corresponding new nodes. I... | 2 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Python || 94.88% Faster || Space is O(1) || 2 Approaches | copy-list-with-random-pointer | 0 | 1 | ```\n#Time Complexity: O(n)\n#Space Complexity: O(n)\nclass Solution1:\n def copyRandomList(self, head: \'Optional[Node]\') -> \'Optional[Node]\':\n prev=dummy=Node(-1)\n curr=head\n d=dict()\n while curr:\n prev.next=Node(-1)\n prev.next.val=curr.val\n d[... | 1 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Python || 94.88% Faster || Space is O(1) || 2 Approaches | copy-list-with-random-pointer | 0 | 1 | ```\n#Time Complexity: O(n)\n#Space Complexity: O(n)\nclass Solution1:\n def copyRandomList(self, head: \'Optional[Node]\') -> \'Optional[Node]\':\n prev=dummy=Node(-1)\n curr=head\n d=dict()\n while curr:\n prev.next=Node(-1)\n prev.next.val=curr.val\n d[... | 1 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Awesome Hashmap Concept | copy-list-with-random-pointer | 0 | 1 | # Using Hashmap--->TC:O(N)\n```\nclass Solution:\n def copyRandomList(self, head: \'Optional[Node]\') -> \'Optional[Node]\':\n dic={None:None}\n cur=head\n while cur:\n dic[cur]=Node(cur.val)\n cur=cur.next\n cur=head\n while cur:\n copy=dic[cur]\n ... | 16 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Awesome Hashmap Concept | copy-list-with-random-pointer | 0 | 1 | # Using Hashmap--->TC:O(N)\n```\nclass Solution:\n def copyRandomList(self, head: \'Optional[Node]\') -> \'Optional[Node]\':\n dic={None:None}\n cur=head\n while cur:\n dic[cur]=Node(cur.val)\n cur=cur.next\n cur=head\n while cur:\n copy=dic[cur]\n ... | 16 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Python3 Solution | copy-list-with-random-pointer | 0 | 1 | \n```\n"""\n# Definition for a Node.\nclass Node:\n def __init__(self, x: int, next: \'Node\' = None, random: \'Node\' = None):\n self.val = int(x)\n self.next = next\n self.random = random\n"""\n\nclass Solution:\n def copyRandomList(self, head: \'Optional[Node]\') -> \'Optional[Node]\':\n ... | 5 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Python3 Solution | copy-list-with-random-pointer | 0 | 1 | \n```\n"""\n# Definition for a Node.\nclass Node:\n def __init__(self, x: int, next: \'Node\' = None, random: \'Node\' = None):\n self.val = int(x)\n self.next = next\n self.random = random\n"""\n\nclass Solution:\n def copyRandomList(self, head: \'Optional[Node]\') -> \'Optional[Node]\':\n ... | 5 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
✔️ [Python3] JUST TWO STEPS ヾ(´▽｀;)ゝ, Explained | copy-list-with-random-pointer | 0 | 1 | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nThe main problem is that a random pointer can point to any node in the list. So we can\'t get by here without a hashmap to remember copied nodes. We need a hashmap that maps the original node to its copy. Having that... | 65 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
✔️ [Python3] JUST TWO STEPS ヾ(´▽｀;)ゝ, Explained | copy-list-with-random-pointer | 0 | 1 | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nThe main problem is that a random pointer can point to any node in the list. So we can\'t get by here without a hashmap to remember copied nodes. We need a hashmap that maps the original node to its copy. Having that... | 65 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Easy Video Solution (Space O(N)->O(1))🔥 || Deep Copy🔥 | copy-list-with-random-pointer | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# ***Detailed and easy Video Solution***\n\nhttps://youtu.be/i5EVm_SfNkA\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n O(N)\n\n- Space complexity:\n<!-- Add your space complexity here, e... | 3 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Easy Video Solution (Space O(N)->O(1))🔥 || Deep Copy🔥 | copy-list-with-random-pointer | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# ***Detailed and easy Video Solution***\n\nhttps://youtu.be/i5EVm_SfNkA\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n O(N)\n\n- Space complexity:\n<!-- Add your space complexity here, e... | 3 | A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.
Construct a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#Deep_copy) of the list. The deep copy should consist of exactly `n` **brand new** nodes, where e... | Just iterate the linked list and create copies of the nodes on the go. Since a node can be referenced from multiple nodes due to the random pointers, make sure you are not making multiple copies of the same node. You may want to use extra space to keep old node ---> new node mapping to prevent creating multiples copies... |
Solution | word-break | 1 | 1 | ```C++ []\nclass Solution {\n public:\n bool wordBreak(string s, vector<string>& wordDict) {\n const int n = s.length();\n const int maxLength = getMaxLength(wordDict);\n const unordered_set<string> wordSet{begin(wordDict), end(wordDict)};\n vector<int> dp(n + 1);\n dp[0] = true;\n\n for (int i = 1; ... | 441 | Given a string `s` and a dictionary of strings `wordDict`, return `true` if `s` can be segmented into a space-separated sequence of one or more dictionary words.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:** s = "leetcode ", wordDict = \[ "... | null |
Python Solution || 100% || 34ms || Beats 98% || | word-break | 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 a string `s` and a dictionary of strings `wordDict`, return `true` if `s` can be segmented into a space-separated sequence of one or more dictionary words.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:** s = "leetcode ", wordDict = \[ "... | null |
Python3 👍||⚡99/83 faster beats 🔥|| clean solution || | word-break | 0 | 1 | \n\n# Complexity\n- Time complexity: O(n*m) \nm is the max length of the words in wordDict \n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity... | 1 | Given a string `s` and a dictionary of strings `wordDict`, return `true` if `s` can be segmented into a space-separated sequence of one or more dictionary words.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:** s = "leetcode ", wordDict = \[ "... | null |
Simple Python solution | word-break | 0 | 1 | \n# Code\n```\nclass Solution:\n def wordBreak(self, s: str, wordDict: List[str]) -> bool:\n dp = [True] + [False] * (len(s))\n for i in range(len(s)):\n for j in range(i, len(s)):\n if s[i:j+1] in wordDict:\n dp[j+1] = dp[i] or dp[j+1]\n return dp[-1... | 1 | Given a string `s` and a dictionary of strings `wordDict`, return `true` if `s` can be segmented into a space-separated sequence of one or more dictionary words.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:** s = "leetcode ", wordDict = \[ "... | null |
✅ 100% DP & DFS [VIDEO] - Segmenting a String | word-break | 1 | 1 | # Intuition\nWhen given a string and a dictionary of words, the problem requires us to determine if the string can be segmented into a sequence of dictionary words. A common way to approach this problem is to use either Dynamic Programming or Depth-First Search with memoization, considering the constraints and properti... | 120 | Given a string `s` and a dictionary of strings `wordDict`, return `true` if `s` can be segmented into a space-separated sequence of one or more dictionary words.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:** s = "leetcode ", wordDict = \[ "... | null |
Ex-Amazon explains a solution with a video, Python, JavaScript, Java and C++ | word-break | 1 | 1 | # Intuition\nUsing dynamic programming to keep a certain substring is true or not.\n\n---\n\n# Solution Video\n## *** Please upvote for this article. *** \n\nhttps://youtu.be/LgoAfakGz5E\n\n# Subscribe to my channel from here. I have 239 videos as of August 4th\nhttp://www.youtube.com/channel/UC9RMNwYTL3SXCP6ShLWVFww?s... | 18 | Given a string `s` and a dictionary of strings `wordDict`, return `true` if `s` can be segmented into a space-separated sequence of one or more dictionary words.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:** s = "leetcode ", wordDict = \[ "... | null |
BFS + Memoization 99.63% Easy to understand | word-break | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFirst thought that comes to mind is prefix, this can be done plainly with BFS, but would lead to memory limit. \n\n```Python []\nclass Solution:\n def wordBreak(self, s: str, wordDict: List[str]) -> bool:\n q = deque([s])\n ... | 5 | Given a string `s` and a dictionary of strings `wordDict`, return `true` if `s` can be segmented into a space-separated sequence of one or more dictionary words.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:** s = "leetcode ", wordDict = \[ "... | null |
EZ 🥧-THON SOLUTION || Explained || (plz help on the complexity part) | word-break | 0 | 1 | # Intuition\nDP will work. First, get a split that works, a.k.a, the word is in `wordDict`. Why not just use recursion? There may be many ways to get to a point. So, the process will be repeated many times. Saving the answer to a position will significantly reduce the runtime.\n\n# Approach\n##### Recursion + Memoizati... | 1 | Given a string `s` and a dictionary of strings `wordDict`, return `true` if `s` can be segmented into a space-separated sequence of one or more dictionary words.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:** s = "leetcode ", wordDict = \[ "... | null |
Easy recursive DFS C++/Python | word-break | 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. $$O(n)$$ --... | 6 | Given a string `s` and a dictionary of strings `wordDict`, return `true` if `s` can be segmented into a space-separated sequence of one or more dictionary words.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:** s = "leetcode ", wordDict = \[ "... | null |
Amortized Linear O(k + m) Solution using Aho-Corasick Automaton | word-break | 0 | 1 | # Approach\nI used an advanced data structure that can detect matches in one-pass. This data structure is called Aho-Corasick automaton. The present solutions in leetcode are at $$O(n^2)$$ solution. This solution runs at $$O(k + m)$$ at test time with an overhead of $$O(n)$$. I hope this helps you impress your future i... | 7 | Given a string `s` and a dictionary of strings `wordDict`, return `true` if `s` can be segmented into a space-separated sequence of one or more dictionary words.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:** s = "leetcode ", wordDict = \[ "... | null |
💯🔥: C++ || Java || Pyhton3 (Space and Time Optimized 💪) | word-break | 1 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo determine if the string s can be segmented into a space-separated sequence of dictionary words, we can use dynamic programming. The idea is to break down the problem into smaller subproblems and build a solution from the subproblems\' ... | 18 | Given a string `s` and a dictionary of strings `wordDict`, return `true` if `s` can be segmented into a space-separated sequence of one or more dictionary words.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:** s = "leetcode ", wordDict = \[ "... | null |
🦀 100% Dynamic Programming | word-break | 0 | 1 | # Intuition\nThe Word Break problem requires us to determine if a given string can be segmented into a sequence of dictionary words. This can be visualized as finding a path through a sequence of characters where each step corresponds to a valid word in the dictionary. Special thanks to vanAmsen for their invaluable in... | 10 | Given a string `s` and a dictionary of strings `wordDict`, return `true` if `s` can be segmented into a space-separated sequence of one or more dictionary words.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:** s = "leetcode ", wordDict = \[ "... | null |
A general template solution for DP Memoization | word-break | 0 | 1 | # Intuition\nThis is exactly the same as [coin change problem](https://leetcode.com/problems/coin-change/) and [Number of Dice Rolls With Target Sum\n](https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/description/). So many medium or hard problems have been solved using this template and now I share i... | 5 | Given a string `s` and a dictionary of strings `wordDict`, return `true` if `s` can be segmented into a space-separated sequence of one or more dictionary words.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:** s = "leetcode ", wordDict = \[ "... | null |
My solution || Memoization || tabulation | word-break | 0 | 1 | \n# Code\n```\nclass Solution:\n\n\n # here start from first if all the prefix word is in dictionary then just call the recursion for the \n # remainning elements\n\n # tabulation\n\n def wordBreak(self, s: str, wordDict: List[str]) -> bool:\n dp = [-1 for i in range(len(s)+1)]\n\n for ind in ... | 1 | Given a string `s` and a dictionary of strings `wordDict`, return `true` if `s` can be segmented into a space-separated sequence of one or more dictionary words.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:** s = "leetcode ", wordDict = \[ "... | null |
Solution | word-break-ii | 1 | 1 | ```C++ []\nclass Solution {\npublic:\n void solve(string s, vector<string>& res, unordered_set<string>& st, vector<string>&temp){\n if(s.length() == 0){\n string str = "";\n for(auto it:temp){\n str += it + " ";\n }\n str.pop_back();\n res.... | 223 | Given a string `s` and a dictionary of strings `wordDict`, add spaces in `s` to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in **any order**.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:... | null |
Python (Simple DFS) | word-break-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. $$O(n)$$ --... | 1 | Given a string `s` and a dictionary of strings `wordDict`, add spaces in `s` to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in **any order**.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:... | null |
recursion + DP always does the job | word-break-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. -->\nrecursive approach\n# Complexity\n- Time complexity:n\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:high\n<!-- Add your space complexity... | 4 | Given a string `s` and a dictionary of strings `wordDict`, add spaces in `s` to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in **any order**.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:... | null |
PYthon basic backtracking solution | word-break-ii | 0 | 1 | \n# Code\n```\nclass Solution:\n def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:\n hmap = {}\n for word in wordDict:hmap[word] = 1\n res = []\n def rec( i , lis ):\n nonlocal res ,s\n if( i == len(s)):\n res.append(" ".join(lis) );\n ... | 2 | Given a string `s` and a dictionary of strings `wordDict`, add spaces in `s` to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in **any order**.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:... | null |
Share my Trie and recursion solutionre | word-break-ii | 0 | 1 | \n# Code\n```\nclass Node:\n def __init__(self, ):\n self.children = {}\n self.is_word = False\n\nclass Trie:\n def __init__(self):\n self.root = Node()\n self.result = []\n\n def insert(self, word: str):\n node = self.root\n for w in word:\n if w not in nod... | 1 | Given a string `s` and a dictionary of strings `wordDict`, add spaces in `s` to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in **any order**.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:... | null |
Backtracking Extreme Simple (Time beats 99.40% | Space beats 82.11%) | word-break-ii | 0 | 1 | # Intuition\nChanged from the solution of Word Break I and adpated backtracking.\n\n# Approach\nJust standard backtracking. Removed cache from Word Break I code.\n\n# Complexity\n- Time complexity:\n$$O(n^2)$$\n\n- Space complexity:\n$$O(n^2)$$\n\n# Code\n```\nclass Solution:\n def wordBreak(self, s: str, wordDict: ... | 1 | Given a string `s` and a dictionary of strings `wordDict`, add spaces in `s` to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in **any order**.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:... | null |
140: Solution with step by step explanation | word-break-ii | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThe time complexity of this algorithm is O(n^3), where n is the length of the input string s. The space complexity is also O(n^3), due to the memoization dictionary. However, in practice, the space complexity will be much lo... | 4 | Given a string `s` and a dictionary of strings `wordDict`, add spaces in `s` to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in **any order**.
**Note** that the same word in the dictionary may be reused multiple times in the segmentation.
**Example 1:**
**Input:... | null |
✅ 99.68% Two-Pointer & Hash Table | linked-list-cycle | 1 | 1 | # Interview Guide - Detecting Cycles in a Linked List: A Dual-Approach Analysis\n\n## Introduction & Problem Understanding\n\nDetecting cycles in a linked list is a classic problem that tests a developer\'s ability to understand and manipulate data structures. In this problem, you are given the head of a singly-linked ... | 122 | Given `head`, the head of a linked list, determine if the linked list has a cycle in it.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is co... | null |
[VIDEO] Visualization of Tortoise and Hare (Floyd's Cycle Detection) Solution | linked-list-cycle | 0 | 1 | https://youtu.be/RRSItF-Ts4Q\n\nThe last pointer in a linked list usually points to a null value to singal the end of the list, but if the last pointer points back to another node in the linked list instead, we now have a cycle. The most common way to solve this problem is by using the Tortoise and Hare algorithm.\n\n... | 7 | Given `head`, the head of a linked list, determine if the linked list has a cycle in it.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is co... | null |
【Video】Solution with two pointers and a bonus video - Python, JavaScript, Java, C++ | linked-list-cycle | 1 | 1 | # Intuition\nThe main idea to solve the question of detecting a cycle in a singly-linked list is to use the concept of two pointers: a "slow" pointer that moves one step at a time and a "fast" pointer that moves two steps at a time. By having these two pointers traverse the list simultaneously, if there is a cycle, the... | 23 | Given `head`, the head of a linked list, determine if the linked list has a cycle in it.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is co... | null |
Using List | linked-list-cycle | 0 | 1 | I just add the each address in the list for every node if an address repeated i.e already existed in the list than I just simply return True . If linked list\'s head reached to the end than return true\n\n# Code\n```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, x):\n# self.... | 1 | Given `head`, the head of a linked list, determine if the linked list has a cycle in it.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is co... | null |
Simple Python solution using slow-fast pointer | linked-list-cycle | 0 | 1 | \n# Code\n```\nclass Solution:\n def hasCycle(self, head: Optional[ListNode]) -> bool:\n slow = fast = head\n while fast and fast.next:\n slow = slow.next\n fast = fast.next.next\n if slow == fast:\n return True\n return False\n``` | 4 | Given `head`, the head of a linked list, determine if the linked list has a cycle in it.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is co... | null |
REALLY EASY 🏀 + 📦 = SLAM DUNK!! | linked-list-cycle | 0 | 1 | \n\n\n\nFirst, you can have a visited dictionary(you have to use defaultdict)\nThen, you kind of go through the list.\nIf you found an already-visited node, return True\notherwise, move pointer forward, visi... | 1 | Given `head`, the head of a linked list, determine if the linked list has a cycle in it.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is co... | null |
Easy Solution | linked-list-cycle | 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 `head`, the head of a linked list, determine if the linked list has a cycle in it.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is co... | null |
using Floyd's Tortoise and Hare algorithm | Simplest Solution | linked-list-cycle | 0 | 1 | # Intuition\nUsing Slow and Fast pointers to detect a cycle.\n\n# Approach\nYou can create a linked list using the ListNode class and pass the head of the list to the hasCycle function. It will return True if there is a cycle in the linked list and False otherwise. The algorithm works by moving two pointers at differen... | 1 | Given `head`, the head of a linked list, determine if the linked list has a cycle in it.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is co... | null |
Python | 2 Pointer | Easy to Understand | linked-list-cycle | 0 | 1 | # Python | 2 Pointer | Easy to Understand\n```\nclass Solution:\n def hasCycle(self, head: ListNode) -> bool:\n if not head:\n return False\n slow = head\n fast = head.next\n while slow != fast:\n if not fast or not fast.next:\n return False\n ... | 4 | Given `head`, the head of a linked list, determine if the linked list has a cycle in it.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is co... | null |
Beats 98%✅ | O( n )✅ | Python (Step by step explanation)✅ | linked-list-cycle | 0 | 1 | # Intuition\nThe problem is to detect whether there is a cycle in a linked list. The intuition is to use two pointers, one slow and one fast, to traverse the linked list. If there is a cycle, the fast pointer will eventually catch up to the slow pointer.\n\n# Approach\n1. Initialize two pointers, `slow` and `fast`, to ... | 2 | Given `head`, the head of a linked list, determine if the linked list has a cycle in it.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is co... | null |
Clean Codes🔥🔥|| Full Explanation✅|| Floyd's Cycle-Finding algorithm✅|| C++|| Java|| Python3 | linked-list-cycle-ii | 1 | 1 | # Intuition :\n- Use a **Floyd\'s Cycle-Finding algorithm** to detect a cycle in a linked list and find the node where the cycle starts.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# What is Floyd\'s Cycle-Finding algorithm ?\n- It is also called **Hare-Tortoise algorithm**\n- The algorithm ... | 400 | Given the `head` of a linked list, return _the node where the cycle begins. If there is no cycle, return_ `null`.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that ta... | null |
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