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Flip bits of the sum of count of set bits of two given numbers

  • Last Updated : 25 May, 2021

Given two numbers A and B, the task is to count the number of set bits in A and B and flip the bits of the obtained sum.

Examples:

Input: A = 5, B = 7
Output: 2
Explanation:
Binary representation of A is 101.
Binary representation of B is 111.
Count of set bits in A and B = 2 + 3 = 5.
Binary representation of the sum obtained = 101
Flipping the bits of the sum, the number obtained is (010)2 = 2. 
Therefore, the required output is 2.

Input: A = 76, B = 35
Output: 1
Explanation: 
Binary representation of A is 1001100
Binary representation of B is 100011
Count of set bits in A and B = 3 + 3 = 6
Binary representation of the sum obtained = 110
Flipping the bits of the sum, the number obtained is (001)2 = 1. 
Therefore, the required output is 1.

 

Naive Approach: The idea to solve this problem is to first traverse through the binary representation of both the numbers and count number of set bits in both the numbers. Finally, add them and invert the bits of the resultant number.



Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count number of
// set bits in integer
int countSetBits(int n)
{
    // Variable for counting set bits
    int count = 0;
    while (n) {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// Function to invert bits of a number
int invertBits(int n)
{
    // Calculate number of bits of N-1;
    int x = log2(n);
 
    int m = 1 << x;
    m = m | m - 1;
    n = n ^ m;
 
    return n;
}
 
// Function to invert the sum
// of set bits in A and B
void invertSum(int A, int B)
{
 
    // Stores sum of set bits
    int temp = countSetBits(A)
               + countSetBits(B);
    cout << invertBits(temp) << endl;
}
 
// Driver Code
int main()
{
    int A = 5;
    int B = 7;
 
    invertSum(A, B);
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
   
class GFG{
  
// Function to count number of
// set bits in integer
static int countSetBits(int n)
{
     
    // Variable for counting set bits
    int count = 0;
     
    while (n != 0)
    {
        n &= (n - 1);
        count++;
    }
    return count;
}
  
// Function to invert bits of a number
static int invertBits(int n)
{
     
    // Calculate number of bits of N-1;
    int x = (int)(Math.log(n) / Math.log(2));
      
    int m = 1 << x;
    m = m | m - 1;
    n = n ^ m;
      
    return n;
}
  
// Function to invert the sum
// of set bits in A and B
static void invertSum(int A, int B)
{
     
    // Stores sum of set bits
    int temp = countSetBits(A) +
               countSetBits(B);
                 
    System.out.print(invertBits(temp));
}
   
// Driver Code
static public void main(String args[])
{
    int A = 5;
    int B = 7;
      
    invertSum(A, B);
}
}
 
// This code is contributed by susmitakundugoaldanga

Python3




# Python3 program for the above approach
import math
 
# Function to count number of
# set bits in integer
def countSetBits(n):
     
    # Variable for counting set bits
    count = 0
 
    while (n != 0):
        n &= (n - 1)
        count += 1
 
    return count
 
# Function to invert bits of a number
def invertBits(n):
     
    # Calculate number of bits of N-1;
    x = (int)(math.log(n) / math.log(2))
 
    m = 1 << x
    m = m | m - 1
    n = n ^ m
 
    return n
 
# Function to invert the sum
# of set bits in A and B
def invertSum(A, B):
     
    # Stores sum of set bits
    temp = countSetBits(A) + countSetBits(B)
 
    print(invertBits(temp))
 
# Driver Code
A = 5
B = 7
 
invertSum(A, B)
 
# This code is contributed by shikhasingrajput

C#




// C# program for the above approach
using System;
 
class GFG{
     
// Function to count number of
// set bits in integer
static int countSetBits(int n)
{
     
    // Variable for counting set bits
    int count = 0;
    while (n != 0)
    {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// Function to invert bits of a number
static int invertBits(int n)
{
     
    // Calculate number of bits of N-1;
    int x = (int)Math.Log(n, 2);
     
    int m = 1 << x;
    m = m | m - 1;
    n = n ^ m;
     
    return n;
}
 
// Function to invert the sum
// of set bits in A and B
static void invertSum(int A, int B)
{
     
    // Stores sum of set bits
    int temp = countSetBits(A) +
               countSetBits(B);
                
    Console.WriteLine(invertBits(temp));
}
 
// Driver Code
static void Main()
{
    int A = 5;
    int B = 7;
     
    invertSum(A, B);
}
}
 
// This code is contributed by divyesh072019

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to count number of
// set bits in integer
function countSetBits(n)
{
     
    // Variable for counting set bits
    var count = 0;
     
    while (n != 0)
    {
        n &= (n - 1);
        count++;
    }
    return count;
}
  
// Function to invert bits of a number
function invertBits(n)
{
     
    // Calculate number of bits of N-1;
    var x = parseInt((Math.log(n) /
                      Math.log(2)));
      
    var m = 1 << x;
    m = m | m - 1;
    n = n ^ m;
      
    return n;
}
  
// Function to invert the sum
// of set bits in A and B
function invertSum(A, B)
{
     
    // Stores sum of set bits
    var temp = countSetBits(A) +
               countSetBits(B);
                 
    document.write(invertBits(temp));
}
   
// Driver Code
var A = 5;
var B = 7;
  
invertSum(A, B);
 
// This code is contributed by Rajput-Ji
 
</script>
Output: 
2

 

Time Complexity: O(logN)
Auxiliary Space: O(1)

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