Given two integers A and B, the task is to count the number of bits needed to be flipped to convert A to B.
Examples:
Input: A = 10, B = 7
Output: 3
binary(10) = 1010
binary(7) = 0111
1010
0111
3 bits need to be flipped.
Input: A = 8, B = 7
Output: 4
Approach: An approach to solve this problem has already been discussed here. Here, the count of bits that need to be flipped can be found by matching all the bits in both the integers one by one. If the bit under consideration differs then increment the count.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the count of bits// to be flipped to convert a to bint countBits(int a, int b){ // To store the required count int count = 0; // Loop until both of them become zero while (a || b) { // Store the last bits in a // as well as b int last_bit_a = a & 1; int last_bit_b = b & 1; // If the current bit is not same // in both the integers if (last_bit_a != last_bit_b) count++; // Right shift both the integers by 1 a = a >> 1; b = b >> 1; } // Return the count return count;}// Driver codeint main(){ int a = 10, b = 7; cout << countBits(a, b); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG {// Function to return the count of bits// to be flipped to convert a to bstatic int countBits(int a, int b){ // To store the required count int count = 0; // Loop until both of them become zero while (a > 0 || b > 0) { // Store the last bits in a // as well as b int last_bit_a = a & 1; int last_bit_b = b & 1; // If the current bit is not same // in both the integers if (last_bit_a != last_bit_b) count++; // Right shift both the integers by 1 a = a >> 1; b = b >> 1; } // Return the count return count;}// Driver codepublic static void main(String[] args) { int a = 10, b = 7; System.out.println(countBits(a, b));}}// This code is contributed by Princi Singh |
Python3
# Python3 implementation of the approach# Function to return the count of bits# to be flipped to convert a to bdef countBits(a, b): # To store the required count count = 0 # Loop until both of them become zero while (a or b): # Store the last bits in a # as well as b last_bit_a = a & 1 last_bit_b = b & 1 # If the current bit is not same # in both the integers if (last_bit_a != last_bit_b): count += 1 # Right shift both the integers by 1 a = a >> 1 b = b >> 1 # Return the count return count# Driver codea = 10b = 7print(countBits(a, b))# This code is contributed by Mohit Kumar |
C#
// C# implementation of the above approachusing System;class GFG {// Function to return the count of bits// to be flipped to convert a to bstatic int countBits(int a, int b){ // To store the required count int count = 0; // Loop until both of them become zero while (a > 0 || b > 0) { // Store the last bits in a // as well as b int last_bit_a = a & 1; int last_bit_b = b & 1; // If the current bit is not same // in both the integers if (last_bit_a != last_bit_b) count++; // Right shift both the integers by 1 a = a >> 1; b = b >> 1; } // Return the count return count;}// Driver codepublic static void Main(String[] args) { int a = 10, b = 7; Console.WriteLine(countBits(a, b));}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript implementation of the approach// Function to return the count of bits// to be flipped to convert a to bfunction countBits(a, b){ // To store the required count var count = 0; // Loop until both of them become zero while (a || b) { // Store the last bits in a // as well as b var last_bit_a = a & 1; var last_bit_b = b & 1; // If the current bit is not same // in both the integers if (last_bit_a != last_bit_b) count++; // Right shift both the integers by 1 a = a >> 1; b = b >> 1; } // Return the count return count;}// Driver code var a = 10, b = 7; document.write(countBits(a, b));</script> |
Output:
3
Time Complexity: O(min(log a, log b))
Auxiliary Space: O(1)
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