Given two numbers A and B. Write a program to count the number of bits needed to be flipped to convert A to B.
Examples:
Input: A = 10, B = 20
Output: 4
Explanation: Binary representation of A is 00001010
Binary representation of B is 00010100
We need to flip highlighted four bits in A to make it B.Input: A = 7, B = 10
Output: 3
Explanation: Binary representation of A is 00000111
Binary representation of B is 00001010
We need to flip highlighted three bits in A to make it B.
Count the number of bits to be flipped to convert A to B using the XOR operator:
To solve the problem follow the below idea:
Calculate (A XOR B), since 0 XOR 1 and 1 XOR 0 is equal to 1. So calculating the number of set bits in A XOR B will give us the count of the number of unmatching bits in A and B, which needs to be flipped
Follow the given steps to solve the problem:
- Calculate the XOR of A and B
- Count the set bits in the above-calculated XOR result
- Return the count
Below is the implementation of the above approach:
C
// C program for the above approach#include <stdio.h>// Function that count set bitsint countSetBits(int n){ int count = 0; while (n > 0) { count++; n &= (n - 1); } return count;}// Function that return count of flipped numberint FlippedCount(int a, int b){ // Return count of set bits in a XOR b return countSetBits(a ^ b);}// Driver codeint main(){ int a = 10; int b = 20; // Function call printf("%d\n", FlippedCount(a, b)); return 0;}// This code is contributed by Sania Kumari Gupta// (kriSania804) |
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function that count set bitsint countSetBits(int n){ int count = 0; while (n > 0) { count++; n &= (n - 1); } return count;}// Function that return count of// flipped numberint FlippedCount(int a, int b){ // Return count of set bits in // a XOR b return countSetBits(a ^ b);}// Driver codeint main(){ int a = 10; int b = 20; // Function call cout << FlippedCount(a, b) << endl; return 0;}// This code is contributed by Sania Kumari Gupta// (kriSania804) |
Java
// Java program for the above approachimport java.util.*;class Count { // Function that count set bits public static int countSetBits(int n) { int count = 0; while (n != 0) { count++; n &= (n - 1); } return count; } // Function that return count of // flipped number public static int FlippedCount(int a, int b) { // Return count of set bits in // a XOR b return countSetBits(a ^ b); } // Driver code public static void main(String[] args) { int a = 10; int b = 20; // Function call System.out.print(FlippedCount(a, b)); }}// This code is contributed by rishabh_jain |
Python3
# Python3 program for the above approach# Function that count set bitsdef countSetBits(n): count = 0 while n: count += 1 n &= (n-1) return count# Function that return count of# flipped numberdef FlippedCount(a, b): # Return count of set bits in # a XOR b return countSetBits(a ^ b)# Driver codeif __name__ == "__main__": a = 10 b = 20 # Function call print(FlippedCount(a, b))# This code is contributed by "Sharad_Bhardwaj". |
C#
// C# program for the above approachusing System;class Count { // Function that count set bits public static int countSetBits(int n) { int count = 0; while (n != 0) { count++; n &= (n - 1); } return count; } // Function that return // count of flipped number public static int FlippedCount(int a, int b) { // Return count of set // bits in a XOR b return countSetBits(a ^ b); } // Driver code public static void Main() { int a = 10; int b = 20; // Function call Console.WriteLine(FlippedCount(a, b)); }}// This code is contributed by vt_m. |
PHP
<?php// php program for the above approach// Function that count set bitsfunction countSetBits($n){ $count = 0; while($n) { $count += 1; $n &= (n-1); } return $count;} // Function that return // count of flipped numberfunction FlippedCount($a, $b){ // Return count of set // bits in a XOR b return countSetBits($a ^ $b);}// Driver code$a = 10;$b = 20;// Function callecho FlippedCount($a, $b);// This code is contributed by mits?> |
Javascript
// Count number of bits to be flipped// to convert A into Bclass Count { // Function that count set bits function countSetBits(n) { var count = 0; while (n != 0) { count++; n &= (n - 1); } return count; } // Function that return count of // flipped number function FlippedCount(a , b) { // Return count of set bits in // a XOR b return countSetBits(a ^ b); } // Driver code var a = 10; var b = 20; document.write(FlippedCount(a, b));// This code is contributed by shikhasingrajput |
Output
4
Time Complexity: O(K) where K is the number of bits
Auxiliary Space: O(1)
Note: Set bits in (a XOR b) can also be computer using built in function __builtin_popcount() in C/C++
Below is the implementation of the above approach:
C++
// C++ program to Count number of bits to be flipped// to convert A into B#include <iostream>using namespace std;// Driver codeint main(){ int a = 10; int b = 20; // Function call cout <<__builtin_popcount(a^b) << endl; return 0;}// This code is contributed by Suruchi Kumari |
C
//C program to Count number of bits to be flipped// to convert A into B#include <stdio.h>// Driver codeint main(){ int a = 10; int b = 20; // Function call printf("%d\n",__builtin_popcount(a^b)); return 0;}// This code is contributed by Suruchi Kumari |
Java
//java codepublic class Main { public static void main(String[] args) { int a = 10; int b = 20; // Function call System.out.println(Integer.bitCount(a ^ b)); }}//code by ksam24000 |
Python3
# Python program to Count number of bits to be flipped# to convert A into B# Driver codeif __name__ == '__main__': a = 10 b = 20 # Function call # Converting int to binary and counting number of bits result = bin(a ^ b).count("1") print(result) |
C#
using System;class Program{ static void Main(string[] args) { int a = 10; int b = 20; // Function call Console.WriteLine(BitCount(a ^ b)); } static int BitCount(int value) { int count = 0; while (value != 0) { count += value & 1; value >>= 1; } return count; }} |
Javascript
// Function to count number of bits to be flipped// to convert A into Bfunction countBits(a, b) { return (a ^ b).toString(2).split('1').length-1;}// Driver codelet a = 10;let b = 20;console.log(countBits(a, b));// This code is contributed by Potta Lokesh |
Output
4
Time Complexity: O(K) where K is the number of bits
Auxiliary Space: O(1)
Count the number of bits to be flipped to convert A to B using the AND operator:
To solve the problem follow the below idea:
Start comparing the bits in A and B, starting from the least significant bit and if (A & 1) is not equal to (B & 1) then the current bit needs to be flipped, as the value of bits is different at this position in both the numbers
Follow the given steps to solve the problem:
- Declare variable flips equal to zero
- Run a loop, while a is greater than zero and b is also greater than zero
- Calculate values of (A AND 1) and (B AND 1)
- If these values are not equal then increase the flip value by 1
- Right shift a and b by 1
- Return flips
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <iostream>using namespace std;int countFlips(int a, int b){ // initially flips is equal to 0 int flips = 0; // & each bits of a && b with 1 // and store them if t1 and t2 // if t1 != t2 then we will flip that bit while (a > 0 || b > 0) { int t1 = (a & 1); int t2 = (b & 1); if (t1 != t2) { flips++; } // right shifting a and b a >>= 1; b >>= 1; } return flips;}int main(){ int a = 10; int b = 20; cout << countFlips(a, b);}// this code is contributed by shivanisinghss2110 |
Java
// Java program for the above approach// CONTRIBUTED BY PRAVEEN VISHWAKARMAimport java.io.*;class GFG { public static int countFlips(int a, int b) { // initially flips is equal to 0 int flips = 0; // & each bits of a && b with 1 // and store them if t1 and t2 // if t1 != t2 then we will flip that bit while (a > 0 || b > 0) { int t1 = (a & 1); int t2 = (b & 1); if (t1 != t2) { flips++; } // right shifting a and b a >>>= 1; b >>>= 1; } return flips; } // Driver code public static void main(String[] args) { int a = 10; int b = 20; // Function call System.out.println(countFlips(a, b)); }} |
Python3
# Python3 program for the above approachdef countFlips(a, b): # initially flips is equal to 0 flips = 0 # & each bits of a && b with 1 # and store them if t1 and t2 # if t1 != t2 then we will flip that bit while(a > 0 or b > 0): t1 = (a & 1) t2 = (b & 1) if(t1 != t2): flips += 1 # right shifting a and b a >>= 1 b >>= 1 return flips# Driver codeif __name__ == "__main__": a = 10 b = 20 # Function call print(countFlips(a, b))# This code is contributed by shivanisinghss2110 |
C#
// C# program for the above approachusing System;class GFG { public static int countFlips(int a, int b) { // initially flips is equal to 0 int flips = 0; // & each bits of a && b with 1 // and store them if t1 and t2 // if t1 != t2 then we will flip that bit while (a > 0 || b > 0) { int t1 = (a & 1); int t2 = (b & 1); if (t1 != t2) { flips++; } // right shifting a and b a >>= 1; b >>= 1; } return flips; } // Driver code public static void Main(String[] args) { int a = 10; int b = 20; // Function call Console.Write(countFlips(a, b)); }}// This code is contributed by shivanisinghss2110 |
Javascript
/*package whatever //do not write package name here */function countFlips(a, b){ // initially flips is equal to 0 var flips = 0; // & each bits of a && b with 1 // and store them if t1 and t2 // if t1 != t2 then we will flip that bit while(a>0 || b>0){ var t1 = (a&1); var t2 = (b&1); if(t1!=t2){ flips++; } // right shifting a and b a>>>=1; b>>>=1; } return flips; } var a = 10; var b = 20; document.write(countFlips(a, b));// This code is contributed by shivanisinghss2110 |
Output
4
Time Complexity: O(K) where K is the number of bits
Auxiliary Space: O(1)
Count the number of bits to be flipped to convert A to B:-
To solve this we just need to do a few simple steps. To know more follow the below steps:-
Approach:
- Convert A and B to binary numbers.
- Compare using ‘equal to’ operator if equal then return 0 otherwise iterate and
- compare the ith of A to ith of B and count the operations
- print the count.
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;string binary(int num){ string str = ""; while (num) { if (num & 1) // 1 str += '1'; else // 0 str += '0'; num >>= 1; // Right Shift by 1 } reverse(str.begin(), str.end()); return str;}int main(){ int a = 10; int b = 20; string astr = binary(a); string bstr = binary(b); // size of the binary strings. int na = astr.size(), nb = bstr.size(); int cnt = 0; // difference between the size of the both a and b // string // int maxi = max(na, nb); int diff = abs(na - nb); // if a size is greater then check it has 1 upto diff // then cnt++; if (na > nb) { for (int i = 0; i < diff; i++) { if (astr[i] == '1') { cnt++; } } } // do the same as above else if (na < nb) { for (int i = 0; i < diff; i++) { if (bstr[i] == '1') { cnt++; } } } na = na - 1; nb = nb - 1; // check from the last if has not equal characters and // cnt++; while (na >= 0 and nb >= 0) { if (astr[na] != bstr[nb]) { cnt++; } na--; nb--; } // print the cnt cout << cnt << endl; return 0;}// this code is contributed by ksam24000 |
Java
import java.util.*;public class Main { public static String binary(int num) { String str = ""; while (num > 0) { if ((num & 1) == 1) // 1 str += '1'; else // 0 str += '0'; num >>= 1; // Right Shift by 1 } return new StringBuilder(str).reverse().toString(); } public static void main(String[] args) { int a = 10; int b = 20; String astr = binary(a); String bstr = binary(b); // size of the binary strings. int na = astr.length(), nb = bstr.length(); int cnt = 0; // difference between the size of the both a and b int diff = Math.abs(na - nb); // if a size is greater then check it has 1 upto diff // then cnt++; if (na > nb) { for (int i = 0; i < diff; i++) { if (astr.charAt(i) == '1') { cnt++; } } } // do the same as above else if (na < nb) { for (int i = 0; i < diff; i++) { if (bstr.charAt(i) == '1') { cnt++; } } } na = na - 1; nb = nb - 1; // check from the last if has not equal characters and // cnt++; while (na >= 0 && nb >= 0) { if (astr.charAt(na) != bstr.charAt(nb)) { cnt++; } na--; nb--; } // print the cnt System.out.println(cnt); }}// This code is contributed by divyansh2212 |
Python3
def binary(num): str = "" while num: if num & 1: # 1 str += '1' else: # 0 str += '0' num >>= 1 # Right Shift by 1 return str[::-1]a = 10b = 20astr = binary(a)bstr = binary(b)# size of the binary strings.na, nb = len(astr), len(bstr)cnt = 0# difference between the size of the both a and b stringdiff = abs(na - nb)# if a size is greater then check it has 1 upto diff then cnt++if na > nb: for i in range(diff): if astr[i] == '1': cnt += 1# do the same as aboveelif na < nb: for i in range(diff): if bstr[i] == '1': cnt += 1na -= 1nb -= 1# check from the last if has not equal characters and cnt++while na >= 0 and nb >= 0: if astr[na] != bstr[nb]: cnt += 1 na -= 1 nb -= 1# print the cntprint(cnt) |
C#
using System;public class Program{ static string Binary(int num) { string str = ""; while (num != 0) { if ((num & 1) == 1) // 1 str += '1'; else // 0 str += '0'; num >>= 1; // Right Shift by 1 } char[] charArray = str.ToCharArray(); Array.Reverse(charArray); return new string(charArray); } public static void Main() { int a = 10; int b = 20; string astr = Binary(a); string bstr = Binary(b); // size of the binary strings. int na = astr.Length, nb = bstr.Length; int cnt = 0; // difference between the size of the both a and b // string int diff = Math.Abs(na - nb); // if a size is greater then check it has 1 upto diff // then cnt++; if (na > nb) { for (int i = 0; i < diff; i++) { if (astr[i] == '1') { cnt++; } } } // do the same as above else if (na < nb) { for (int i = 0; i < diff; i++) { if (bstr[i] == '1') { cnt++; } } } na = na - 1; nb = nb - 1; // check from the last if has not equal characters and // cnt++; while (na >= 0 && nb >= 0) { if (astr[na] != bstr[nb]) { cnt++; } na--; nb--; } // print the cnt Console.WriteLine(cnt); }}// ksam24000 |
Javascript
function binary(num) { let str = ''; while (num) { if (num & 1) str += '1'; else str += '0'; num >>= 1; } return str.split('').reverse().join('');}let a = 10;let b = 20;let astr = binary(a);let bstr = binary(b);let na = astr.length;let nb = bstr.length;let cnt = 0;let diff = Math.abs(na - nb);if (na > nb) { for (let i = 0; i < diff; i++) { if (astr[i] === '1') cnt++; }} else if (na < nb) { for (let i = 0; i < diff; i++) { if (bstr[i] === '1') cnt++; }}na--;nb--;while (na >= 0 && nb >= 0) { if (astr[na] !== bstr[nb]) cnt++; na--; nb--;}console.log(cnt);// THIS CODE IS CONTRIBUTED BY CHANDAN AGARWAL |
Output
4
Time complexity- O(log N)
Auxiliary Space – O(N)
Thanks to Sahil Rajput for providing the above implementation.
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