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leetcode
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‎README.md

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@@ -50,6 +50,8 @@ This repo contain leetcode solution using DART and GO programming language. Most
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- [Best Time to Buy and Sell Stock](BestTimeToBuyAndSellStock/best_time_to_buy_and_sell_stock.dart)
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- [Valid Palindrome](ValidPalindrome/valid_palindrome.dart)
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- [Push Dominoes](PushDominoes/push_dominoes.dart)
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- [Single Number](SingleNumber/single_number.dart)
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- [Remove Nth Node From End of List](RemoveNthNodeFromEndOfList/remove_nth_node_from_end_of_list.dart)
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## Reach me via
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‎RemoveNthNodeFromEndOfList/remove_nth_node_from_end_of_list.dart

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/*
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-* Remove Nth Node From End of List *-
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Given the head of a linked list, remove the nth node from the end of the list and return its head.
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Example 1:
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Input: head = [1,2,3,4,5], n = 2
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Output: [1,2,3,5]
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Example 2:
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Input: head = [1], n = 1
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Output: []
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Example 3:
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Input: head = [1,2], n = 1
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Output: [1]
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Constraints:
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The number of nodes in the list is sz.
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1 <= sz <= 30
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0 <= Node.val <= 100
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1 <= n <= sz
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*/
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// Definition for singly-linked list.
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class ListNode {
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int val;
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ListNode? next;
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ListNode([this.val = 0, this.next]);
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}
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// Two Pointers
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class Solution {
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// Runtime: 465 ms, faster than 52.94% of Dart online submissions for Remove Nth Node From End of List.
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// Memory Usage: 142.4 MB, less than 88.24% of Dart online submissions for Remove Nth Node From End of List.
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ListNode? removeNthFromEnd(ListNode? head, int n) {
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// if the head is null we will simply return null
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if (head == null) return null;
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// Two pointers of our linked list
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ListNode? fast = head;
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ListNode? slow = head;
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// looping through the value and pointing it to the next
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for (int i = 1; i <= n; ++i) fast = fast!.next;
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// if the previous value is null but next is not
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ListNode? prev = null;
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// while the first pointer is not null
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while (fast != null) {
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// fast will hold the next value
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fast = fast.next;
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// previous will hold the second pointer
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prev = slow;
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slow = slow!.next;
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}
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// if previous is null
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if (prev == null) {
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// head will pint to the next value
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head = head.next;
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} else {
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prev.next = slow!.next;
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slow.next = null;
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}
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// else we will simply return head
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return head;
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}
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}
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// Three Pointer
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/*
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Approach :- First increase the fast pointer by n steps to ensure that when the fast pointer points to null , at the same time, the slow pointer points to the location of the node which is to be removed
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fast pointer : to move n steps ahead of slow pointer
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slow pointer : this will point to the exact node which is to be removed once the fast pointer reaches null
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prev pointer: this will point to the node before the slow pointer because this pointer will bypass the node which is to be removed and get connected directly to the node next to slow.
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*/
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class B {
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// Runtime: 509 ms, faster than 50.00% of Dart online submissions for Remove Nth Node From End of List.
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// Memory Usage: 142.5 MB, less than 79.41% of Dart online submissions for Remove Nth Node From End of List.
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ListNode? removeNthFromEnd(ListNode? head, int n) {
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ListNode? fast = head;
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ListNode? slow = head;
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ListNode? prev = null;
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//Loop to take the fast pointer n steps ahead of the slow pointer.
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while (n > 0) {
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fast = fast?.next;
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n--;
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}
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//If fast pointer points to null, it means that the node has to be removed from the beginning.
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if (fast == null) return slow?.next;
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while (fast != null) {
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fast = fast.next;
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prev = slow;
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slow = slow?.next;
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}
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if (prev != null)
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prev.next = slow?.next;
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else if (prev == null) return null;
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return head;
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}
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}
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class C {
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// Runtime: 488 ms, faster than 52.94% of Dart online submissions for Remove Nth Node From End of List.
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// Memory Usage: 142.4 MB, less than 79.41% of Dart online submissions for Remove Nth Node From End of List.
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ListNode? removeNthFromEnd(ListNode? head, int n) {
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ListNode? fast = head, slow = head;
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for (int i = 1; i <= n; i++) fast = fast?.next;
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// for cases where n = length of linked-list i.e. if first node is to be deleted
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if (fast == null) {
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head = head?.next;
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return head;
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}
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while (fast?.next != null) {
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slow = slow?.next;
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fast = fast?.next;
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}
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slow?.next = slow.next?.next;
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return head;
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}
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}

‎RemoveNthNodeFromEndOfList/remove_nth_node_from_end_of_list.go

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package main
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// Definition for singly-linked list.
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type ListNode struct {
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Val int
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Next *ListNode
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}
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func removeNthFromEnd(head *ListNode, n int) *ListNode {
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dummy := new(ListNode)
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dummy.Next = head
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fast := head
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slow := dummy
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for i := 0; i < n; i++ {
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fast = fast.Next
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}
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for fast != nil {
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fast = fast.Next
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slow = slow.Next
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}
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slow.Next = slow.Next.Next
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return dummy.Next
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}

‎RemoveNthNodeFromEndOfList/remove_nth_node_from_end_of_list.md

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# 🔥 DART || 3 Solutions || With explanation
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## Definition for singly-linked list
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```dart
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class ListNode {
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int val;
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ListNode? next;
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ListNode([this.val = 0, this.next]);
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}
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```
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## Solution - 1 Two-Pointers
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```dart
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class Solution {
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// Runtime: 465 ms, faster than 52.94% of Dart online submissions for Remove Nth Node From End of List.
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// Memory Usage: 142.4 MB, less than 88.24% of Dart online submissions for Remove Nth Node From End of List.
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ListNode? removeNthFromEnd(ListNode? head, int n) {
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// if the head is null we will simply return null
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if (head == null) return null;
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// Two pointers of our linked list
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ListNode? fast = head;
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ListNode? slow = head;
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// looping through the value and pointing it to the next
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for (int i = 1; i <= n; ++i) fast = fast!.next;
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// if the previous value is null but next is not
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ListNode? prev = null;
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// while the first pointer is not null
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while (fast != null) {
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// fast will hold the next value
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fast = fast.next;
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// previous will hold the second pointer
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prev = slow;
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slow = slow!.next;
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}
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// if previous is null
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if (prev == null) {
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// head will pint to the next value
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head = head.next;
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} else {
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prev.next = slow!.next;
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slow.next = null;
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}
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// else we will simply return head
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return head;
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}
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}
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```
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## Solution - 2 Three Pointers
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Approach :- First increase the fast pointer by n steps to ensure that when the fast pointer points to null , at the same time, the slow pointer points to the location of the node which is to be removed
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fast pointer : to move n steps ahead of slow pointer
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slow pointer : this will point to the exact node which is to be removed once the fast pointer reaches null
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prev pointer: this will point to the node before the slow pointer because this pointer will bypass the node which is to be removed and get connected directly to the node next to slow.
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```dart
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class Solution {
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// Runtime: 509 ms, faster than 50.00% of Dart online submissions for Remove Nth Node From End of List.
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// Memory Usage: 142.5 MB, less than 79.41% of Dart online submissions for Remove Nth Node From End of List.
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ListNode? removeNthFromEnd(ListNode? head, int n) {
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ListNode? fast = head;
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ListNode? slow = head;
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ListNode? prev = null;
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//Loop to take the fast pointer n steps ahead of the slow pointer.
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while (n > 0) {
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fast = fast?.next;
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n--;
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}
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//If fast pointer points to null, it means that the node has to be removed from the beginning.
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if (fast == null) return slow?.next;
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while (fast != null) {
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fast = fast.next;
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prev = slow;
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slow = slow?.next;
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}
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if (prev != null)
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prev.next = slow?.next;
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else if (prev == null) return null;
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return head;
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}
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}
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```
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## Solution - 3 Without Using Dummy Node
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```dart
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class Solution {
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// Runtime: 488 ms, faster than 52.94% of Dart online submissions for Remove Nth Node From End of List.
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// Memory Usage: 142.4 MB, less than 79.41% of Dart online submissions for Remove Nth Node From End of List.
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ListNode? removeNthFromEnd(ListNode? head, int n) {
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ListNode? fast = head, slow = head;
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for (int i = 1; i <= n; i++) fast = fast?.next;
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// for cases where n = length of linked-list i.e. if first node is to be deleted
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if (fast == null) {
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head = head?.next;
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return head;
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}
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while (fast?.next != null) {
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slow = slow?.next;
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fast = fast?.next;
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}
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slow?.next = slow.next?.next;
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return head;
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}
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}
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```
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## Bonus Solution - Golang
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```go
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// Runtime: 0 ms, faster than 100.00% of Go online submissions for Remove Nth Node From End of List.
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// Memory Usage: 2.3 MB, less than 31.13% of Go online submissions for Remove Nth Node From End of List.
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func removeNthFromEnd(head *ListNode, n int) *ListNode {
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dummy := new(ListNode)
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dummy.Next = head
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fast := head
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slow := dummy
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for i := 0; i < n; i++ {
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fast = fast.Next
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}
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for fast != nil {
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fast = fast.Next
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slow = slow.Next
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}
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slow.Next = slow.Next.Next
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return dummy.Next
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}
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```

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