Hamming Distance

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, calculate the Hamming distance.

Note:
0 ≤ x, y < 231.

Example: 

Input: x = 1, y = 4

Output: 2

Explanation:
1   (0 0 0 1)
4   (0 1 0 0)
       ↑   ↑

The above arrows point to positions where the corresponding bits are different.

 class Solution {
    public int hammingDistance(int x, int y) {
        String str = Integer.toBinaryString(x^y);
        return str.replaceAll("0","").length();
    }
}


Here, we are going to solve in more easiest way. The hint hidden in this problem is, we need to convert integer to binary and compare both for any differences in 1's position.

1) Do XOR for both numbers.(XOR of same digits (0, 1) will be same). From this, we get difference in 1's position, but it will be in int type.
2) Convert int to binary using toBinaryString().
3) Replace all 0's and length after replacing will be the output.

Related Post:

Click here for April month challenges with solution and explanation

Click Follow button to receive latest updates from us instantly.

Please let us know about your views in comment section.

Help Others, Please Share..!!!!
                                                     😍..Happy Coding...😍
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!

Comments

  1. Unknown5 July 2020 at 18:35

    why are we doing x^y?

    and whats wrong with this?
    public int hammingDistance(int x, int y) {

    int count=0;
    String a= Integer.toBinaryString(x);
    String b= Integer.toBinaryString(y);

    for(int i=1, j=1;i<a.length() && j<b.length();++i,++j){
    if(a.charAt(i)!=b.charAt(j)){
    count++;
    }
    }

    return count;
    }

    ReplyDelete
  2. Fellow Programmer6 July 2020 at 12:47

    Hi,
    Thanks for your valuable question. Reason for doing x^y is given below solution. Still having doubt on use of XOR, check out XOR table in this page and compare with explanation https://en.wikipedia.org/wiki/XOR_gate .
    Also, Integer.toBinaryString() converts int to binary string, but it will automatically all 0's in LHS
    eg) x =1, y = 4 According to your code, it will become a =1 , b = 101 (In a, it eliminates all 0's before 1 but we need 001 for comparison with b = 101). Thus, your code fails in for loop. Hope it clears your doubt. Follow us to receive daily updates from us instantly.

    ReplyDelete

Post a Comment

Popular posts from this blog

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.substring(i+1);             }             if(!str.contains(s.charAt(i)+""))                 return i;         }         return -1;     } } Try it on Leetcode Here, we are going to use substring method. 1) If character is at end, then take substring from 0 till before last character of string. 2) Else make a temporary string, by removing exactly that particular character using index. [e.g..
Read more

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 val, TreeNode left, TreeNode right) {  *         this.val = val;  *         this.left = left;  *         this.right = right;  *     }
Read more

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 repeatLen
Read more

Smallest Range II

Given an array   A   of integers, for each integer   A[i]   we need to choose   either  x = -K  or  x = K , and add   x   to   A[i]  (only once) . After this process, we have some array  B . Return the smallest possible difference between the maximum value of  B  and the minimum value of  B . Example 1: Input: A = [1] , K = 0 Output: 0 Explanation : B = [1] Example 2: Input: A = [0,10] , K = 2 Output: 6 Explanation : B = [2,8] Example 3: Input: A = [1,3,6] , K = 3 Output: 3 Explanation : B = [4,6,3] Note: 1 <= A.length <= 10000 0 <= A[i] <= 10000 0 <= K <= 10000 Solution: class Solution {     public int smallestRangeII(int[] A, int K) {         int i, min=0, max=0, min1, min2, max1, max2, length = A.length, diff
Read more