Split Linked List in Parts | Clean | Simple Code | Dry Run | Leetcode - 725 | codestorywithMIK скачать в хорошем качестве

Split Linked List in Parts | Clean | Simple Code | Dry Run | Leetcode - 725 | codestorywithMIK

This is the 16th Video on our Linked List Playlist. In this video we will try to solve a very very famous Linked List Problem - Split Linked List in Parts | Clean | Simple Code | Dry Run | Leetcode - 725 | codestorywithMIK Trust me, this will no longer be a Medium Problem. I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY. We will do live coding after explanation and see if we are able to pass all the test cases. Problem Name : Split Linked List in Parts | Clean | Simple Code | Dry Run | Leetcode - 725 | codestorywithMIK Company Tags : AMAZON My solutions on Github : https://github.com/MAZHARMIK/Intervie... Leetcode Link : https://leetcode.com/problems/split-l... My DP Concepts Playlist :    • Roadmap for DP | How to Start DP ? | Topic...   My Graph Concepts Playlist :    • Graph Concepts & Qns - 1 : Graph will no m...   My GitHub Repo for interview preparation : https://github.com/MAZHARMIK/Intervie... Subscribe to my channel :    / @codestorywithmik   Instagram :   / codestorywithmik   Facebook :   / 100090524295846   Twitter :   / cswithmik   ╔═╦╗╔╦╗╔═╦═╦╦╦╦╗╔═╗ ║╚╣║║║╚╣╚╣╔╣╔╣║╚╣═╣ ╠╗║╚╝║║╠╗║╚╣║║║║║═╣ ╚═╩══╩═╩═╩═╩╝╚╩═╩═╝ Summary : The approach splits a singly linked list into `k` parts as evenly as possible. Here's the step-by-step breakdown: 1. **Calculate the Length**: First, traverse the linked list to determine its total length `L`. 2. **Determine the Size of Each Part**: Calculate `eachBucketNodes` as the base size of each part by dividing `L` by `k`. Calculate `remainderNodes` as the remainder of `L` divided by `k`, which tells us how many parts should have one extra node. 3. **Split the List**: Create an array of size `k` to store the head of each part. Traverse the list again, and for each part, assign nodes to it. Each part gets `eachBucketNodes` nodes, and the first `remainderNodes` parts receive one extra node. After assigning the nodes for each part, disconnect it from the next part by setting the `next` pointer of the last node to `null`. 4. **Handle Remaining Parts**: If there are more parts than nodes, the remaining parts in the array are set to `null`. This approach ensures that the list is split as evenly as possible, with the extra nodes distributed to the first parts. #coding #helpajobseeker #easyrecipes #leetcode #leetcodequestionandanswers #leetcodesolution #leetcodedailychallenge #leetcodequestions #leetcodechallenge #hindi #india #coding #helpajobseeker #easyrecipes #leetcode #leetcodequestionandanswers #leetcodesolution #leetcodedailychallenge#leetcodequestions #leetcodechallenge #hindi #india #hindiexplanation #hindiexplained #easyexplaination #interview#interviewtips #interviewpreparation #interview_ds_algo #hinglish #github #design #data #google #video #instagram #facebook #CodeStoriWithMIK #CodeStryWithMIK #C0deStoryWithMIK #KodeStoryWithMIK #Cod3StoryWithMIK #CodeSoryWithMIK #CodeStorryWithMIK #CoedStoryWithMIK #CodStoryWitMIK #C0deSt0ryWitMIK #CodeWithMik #KodeWithMik #C0deWithMik #CodeWitMik #Cod3WithMik #CodeWitMIK #CodeWidMik #CodeWihMik #C0deWitMik #KodeWitMik