# Minimum number of elements to be removed to make XOR maximum using C++.

## Problem statement

Given a number N. The task is to find the minimum number of elements to be removed in between to N such that the XOR obtained from the remaining elements is maximum.

## Algorithm

1. If n is 1 or 2 then there is no need to remove any element. Hence answer is zero
2. Find a number which is power of 2 and greater than or equal to. Let us call this number as nextNumber
2.1. If n == nextNumber or n == (nextNumber – 1) then answer is 1
2.2. If n = (nextNumber -2) then answer is 0
3. If n is an even then answer is 1 otherwise 2

## Example

#include <iostream>
using namespace std;
int nextPowerOf2(int n){
if (n && !(n & (n - 1))) {
return n;
}
int cnt = 0;
while (n) {
n = n / 2;
++cnt;
}
return (1 << cnt);
}
int elmentsToBeRemoved(int n){
if (n == 1 || n == 2) {
return 0;
}
int nextNumber = nextPowerOf2(n);
if (n == nextNumber || n == nextNumber -1) {
return 1;
} else if (n == nextNumber - 2) {
return 0;
} else if (n & 1) {
return 2;
} else {
return 1;
}
}
int main(){
int n = 10;
cout << "Numbers to be removed = " <<
elmentsToBeRemoved(n) << endl;
return 0;
}

## Output

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

Numbers to be removed = 1s

Updated on: 31-Oct-2019

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