Given two strings S and T, determine if they are both one edit distance apart. Java Solution public boolean isOneEditDistance(String s, String t) { if(s==null || t==null) return false; int m = s.length(); int n = t.length(); if(Math.abs(m-n)>1){ return false; } int i=0; int j=0; int count=0; while(i<m&&j<n){ if(s.charAt(i)==t.charAt(j)){ i++; j++; … Read more
Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, …) which sum to n. For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9. Java Solution This … Read more
Reverse a singly linked list. Java Solution 1 – Iterative public ListNode reverseList(ListNode head) { if(head==null||head.next==null) return head; ListNode p1 = head; ListNode p2 = p1.next; head.next = null; while(p1!=null&& p2!=null){ ListNode t = p2.next; p2.next = p1; p1 = p2; p2 = t; } return p1; }public ListNode reverseList(ListNode head) { … Read more
Given two strings s and t, determine if they are isomorphic. Two strings are isomorphic if the characters in s can be replaced to get t. For example,”egg” and “add” are isomorphic, “foo” and “bar” are not. Java Solution 1 We can define a map which tracks the char-char mappings. If a value is already … Read more
There are a total of n courses you have to take, labeled from 0 to n – 1. Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]. Given the total number of courses and a list of prerequisite pairs, … Read more
Given a board with m by n cells, each cell has an initial state live (1) or dead (0). Each cell interacts with its eight neighbors (horizontal, vertical, diagonal) using the following four rules: Any live cell with fewer than two live neighbors dies, as if caused by under-population. Any live cell with two or … Read more
Given an array of citations (each citation is a non-negative integer) of a researcher, write a function to compute the researcher’s h-index. A scientist has index h if h of his/her N papers have at least h citations each, and the other N − h papers have no more than h citations each. For example, … Read more
You are playing the following Bulls and Cows game with your friend: You write down a number and ask your friend to guess what the number is. Each time your friend makes a guess, you provide a hint that indicates how many digits in said guess match your secret number exactly in both digit and … Read more
Implement a trie with insert, search, and startsWith methods. Java Solution 1 A trie node should contains the character, its children and the flag that marks if it is a leaf node. You can use the trie in the following diagram to walk though the Java solution. class TrieNode { char c; HashMap<Character, TrieNode> children … Read more
Given an array containing n distinct numbers taken from 0, 1, 2, …, n, find the one that is missing from the array. For example, given nums = [0, 1, 3] return 2. Java Solution 1 – Math public int missingNumber(int[] nums) { int sum=0; for(int i=0; i<nums.length; i++){ sum+=nums[i]; } int n=nums.length; return … Read more
Java Solution This problem can be solved by BFS. We define one matrix for tracking the distance from each building, and another matrix for tracking the number of buildings which can be reached. public int shortestDistance(int[][] grid) { int[][] distance = new int[grid.length][grid[0].length]; int[][] reach = new int[grid.length][grid[0].length]; int numBuilding = 0; for … Read more
Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Java Solution Inorder traveral will return values in an increasing order. So if an element is less than its previous element,the previous element is a swapped node. public class Solution { TreeNode first; TreeNode second; TreeNode … Read more
Given n nodes labeled from 0 to n – 1 and a list of undirected edges (each edge is a pair of nodes), check if these edges form a valid tree. Analysis This problem can be converted to finding the cycle from a graph. It can be solved by using DFS (Recursion) or BFS (Queue). … Read more
Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. Java Solution public int maxDepth(TreeNode root) { if(root==null) return 0; int leftDepth = maxDepth(root.left); int rightDepth = maxDepth(root.right); int bigger = Math.max(leftDepth, rightDepth); … Read more
Given a binary tree, return all root-to-leaf paths. 1. Naive DFS Solution A typical depth-first search problem. public List<String> binaryTreePaths(TreeNode root) { ArrayList<String> finalResult = new ArrayList<String>(); if(root==null) return finalResult; ArrayList<String> curr = new ArrayList<String>(); ArrayList<ArrayList<String>> results = new ArrayList<ArrayList<String>>(); dfs(root, results, curr); for(ArrayList<String> al : results){ StringBuilder sb = new … Read more