HDegano

//Assumes there are no duplicate 
//O(log n)
public int FindMinRotatedArray(int[] ar){
  
  int n = ar.Length;
  int lo = 0;
  int hi = n - 1;
  while(lo <= hi){
    
    if(ar[lo] <= ar[hi]) // (sub) array is already sorted, yay!

HDegano

/*
Assumes no duplicate
O(logN)
*/

public int FindElementInRotatedSortedArray(int[] ar, int k){
  
  int n = ar.Length;
  int lo = 0;
  int hi = n - 1;

HDegano

public static bool balanced_brackets(string S)
{
    var open = new List<char>(){'(', '{', '['};

    var stack = new Stack<char>();

    foreach (var i in S)
    {
        if (open.Contains(i)) stack.Push(i);
        

HDegano

public class MinStack {
    
    private Stack<int> _stack = new Stack<int>();
    private Stack<int> _min = new Stack<int>();
    
    public void Push(int x) {
        _stack.Push(x);    
        
        if(_min.Count == 0) _min.Push(x);
        else {

HDegano

    public class TreeNode {
        public int val;
        public TreeNode left;
        public TreeNode right;
        public TreeNode(int x) { val = x; }
    }
    
    public bool IsBalanced(TreeNode root) {
        if(root == null) return true;
        

HDegano

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode MergeTwoLists(ListNode l1, ListNode l2) {

HDegano

/**
 * Definition for binary tree
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {

HDegano

/*
Requires O(mn) space
*/
public class Solution {
    public int MinDistance(string word1, string word2) {
        
            int[,] dp = new int[word2.Length + 1, word1.Length + 1];

            for (int i = 0; i <= word1.Length; i++) dp[0, i] = i;
            for (int i = 0; i <= word2.Length; i++) dp[i, 0] = i;

HDegano

public class Solution{
  
  public ListNode FindKthNode(ListNode head, int n){
    
    if(head == null || n <= 0) return null;
    
    ListNode fast = head;

    //Verify what is Kth to the last
    // 1 -> 2 -> 3 and k = 1, if K = 3 or K = 2. For now assume the former is correct

HDegano

public class Solution{
  
  public ListNode ReverseLinkList(ListNode head){
    
    if(head == null) return null;
    
    ListNode current = head;
    ListNode prev = null;
    ListNode reverseHead = null;