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Java Solution
The intention behind this blog is to help all my fellow programmers. Also to provide solutions for various coding platform problems with solution and explanation. I will add many tutorials in future. I am not a professional blogger but whenever time permits I will post. I hope you’ll enjoy the content and find it useful! Check out my Writer blog for a more valuable quotes. For any queries or suggestions, please feel free to reach out to me. Happy coding!

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First Unique Character in a String

Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1. Examples: s = "leetcode" return 0. s = "loveleetcode", return 2. Note: You may assume the string contain only lowercase letters. class Solution {     public int firstUniqChar(String s) {         int length = s.length();         String str;         for(int i=0;i<length;i++){             if(i==length-1){                 str = s.substring(0, i);             }             else{                 str = s.substring(0,i)+s.sub...
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Balanced Binary Tree

Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as: a binary tree in which the left and right subtrees of  every  node differ in height by no more than 1. Example 1: Input: root = [3,9,20,null,null,15,7] Output: true Example 2: Input: root = [1,2,2,3,3,null,null,4,4] Output: false Example 3: Input: root = [] Output: true   Constraints: The number of nodes in the tree is in the range  [0, 5000] . -10 4  <= Node.val <= 10 4 Solution: /**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode() {}  *     TreeNode(int val) { this.val = val; }  *     TreeNode(int ...
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Majority Element

Given an array of size n , find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the majority element always exist in the array. Example 1: Input: [3,2,3] Output: 3 Example 2: Input: [2,2,1,1,1,2,2] Output: 2   class Solution { public int majorityElement(int[] nums) { Arrays.sort(nums); return nums[nums.length/2]; } } Here, we are sorting the entire array using sort() of Arrays class. Then, number at index position n/2 would be a majority element. There is no chance of wrong values in odd or even length.   Below is another method using HashMap. METHOD 2: class Solution { public int majorityElement(int[] nums) { HashMap<Integer, Integer>list = new HashMap<Integer, Integer>(); int ...
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Sequential Digits

An integer has sequential digits if and only if each digit in the number is one more than the previous digit. Return a sorted list of all the integers in the range [low, high]  inclusive that have sequential digits. Example 1: Input: low = 100, high = 300 Output: [123,234] Example 2: Input: low = 1000, high = 13000 Output: [1234,2345,3456,4567,5678,6789,12345] Constraints: 10 <= low <= high <= 10^9 class Solution {     public List<Integer> sequentialDigits(int low, int high) {         List<Integer> result = new ArrayList();         String str = "123456789";         int i, j, length = str.length(), num;         for( i = 2; i <= length; i++) {             for( j = 0; j <= length - i; j++) {      ...
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