[Question]: Given a singly linked list head sort it with with all with zeros, ones and two’sExample:Input: 1 -> 1 -> 2 -> 0 -> 2 -> 0 -> 1 -> NULL Output: 0 -> 0 -> 1 -> 1 -> 1 -> 2 -> 2 -> NULL Approach: Create count array which stores count of 0’s, 1’s […]

[Question]: Given Head of singly linked list return a sorted linkedlist head(ascending order.)Criteria: -TC should O(n) SC: O(1) Input: head = [4,2,1,3] Output: [1,2,3,4] #Approach : We can use merge sort: which work on divide & conquer

[Question]: Input singly linked list return true if linked list is palindrome.Example: – Input: head = [1,2,2,1] Output: true Approach# Reverse Linked list & compare first and last item

[Question] Given a Doubly Linked List, the task is to reverse the given Doubly Linked List. Here we can can solve by two approach #1 Iterative #2 Recursive For information about usage visit my Githhub Page

[Question]: Given the heads of two singly linked-lists headA and headB, return the node at which the two lists intersect. If the two linked lists have no intersection at all, return null.

[Question] Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list. The first node is considered odd, and the second node is even, and so on. You must solve the problem in O(1) extra space complexity and O(n) time complexity. Example 1: Input: head = [1,2,3,4,5] Output: [1,3,5,2,4]

[Question] Reverse Singly Link List using Iterative & Recursion approach.

[Question]: Given the head of a linked list, remove the nth node from the end of the list and return its head. Input: head = [1,2,3,4,5], n = 4 Output: [1,3,4,5] #Approach Two pointer Step: 1 Move fast pointer till countStep: 2 Increment fast pointer till we get fast?.next == nil if we got nil just remove mid pointer…

Here is example of Doubly Link list in swift

[Question]: You are given the head of a linked list. Delete the middle node, and return the head of the modified linked list. The middle node of a linked list of size n is the ⌊n / 2⌋th node from the start using 0-based indexing, where ⌊x⌋ denotes the largest integer less than or equal to x.