# Minimum flips required to maximize a number with k set bits in C++.

## Problem statement

Given two numbers n and k, we need to find the minimum number of flips required to maximize given number by flipping its bits such that the resulting number has exactly k set bits. Please note input must satify condition that k < number of bits in n.

Example

• Let us suppose n = 9 and k = 2

• Binary representation of 9 is − 1001. It contains 4 bits.

• Largest 4 digit binary number with 2 set bits is − 1100 i.e. 12

• To convert 1001 to 1100 we have to flip highlighed 2 bits

## Algorithm

1. Count the number of bits in n. Let us call this variable as bitCount. we can use log2(n) function from math library as fllows:
bitCount = log2(n) + 1;
2. Find largest number with k set bits as follows: maxNum = pow(2, k) - 1;
3. Find the largest number possible with k set bits and having exactly same number of bits as n as follows:
maxNum = maxNum << (bitCount - k);
4. Calculate (n ^ maxNum) and count number of set bits.

## Example

#include <iostream>
#include <cmath>
using namespace std;
int getSetBits(int n){
int cnt = 0;
while (n) {
++cnt;
n = n & (n - 1);
}
return cnt;
}
int minFlipsRequired(int n, int k){
int bitCount, maxNum, flipCount;
bitCount = log2(n) + 1;
maxNum = pow(2, k) - 1;
maxNum = maxNum << (bitCount - k);
flipCount = n ^ maxNum;
return getSetBits(flipCount);
}
int main(){
cout << "Minimum required flips: " << minFlipsRequired(9, 2) << "\n";
return 0;
}

## Output

When you compile and execute above program. It generates following output −

Minimum required flips: 2