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# Equal Sum and XOR in C++

In this problem, we are given an integer n. Our task is to create a program to find the count of integers from i = 0 to n, where sum is equal to XOR i.e. *(n+i) = (n^i).*

## Let’s take an example to understand the problem,

**Input: ** n = 4

**Output: **4

**Explanation: **

Considering all values of i from 0 to n,

i = 0, 4 + 0 = 4, 4^0 = 4

i = 1, 4 + 1 = 5, 4^1 = 5

i = 2, 4 + 2 = 6, 4^2 = 6

i = 3, 4 + 3 = 7, 4^3 = 7

i = 4, 4 + 4 = 8, 4^4 = 0

Count = 4

## Solution Approach:

A simple solution is to find the values of the sum of n and i and xor of n and i. Compare both these values and then count the values for which they are equal.

## Algorithm:

**Step 1: **Loop for all values from i = 0 to n.

**Step 1.1: **Find the value of (n + i).

**Step 1.2: **Find the value of (n^i).

**Step 1.3: **compare values found in step 1.1 and 1.2.**Step 1.4: **If they are equal, increase count.

**Step 2: **Print *count *values.

## Program to illustrate the working of our solution,

## Example

#include <iostream> using namespace std; int main() { int n = 5; int counter = 0; for(int i=0; i<=n; i++ ) if ( (n+i) == (n^i) ) counter++; cout<<"The count of integers with equal sum and XOR is "<<counter; return 0; }

## Output −

The count of integers with equal sum and XOR is 2

The method is good but their can be a better solution to the problem, which is using the fact that

If ** n^i = n+i**, then

**.**

*n&i = 0*If the value of n&i = 0, for that we need the two numbers to have opposite set and unset bits. And we need to count such values. Here is a program to do it,

## Example

#include <iostream> using namespace std; int countValuesWithEqualSumXOR(int n) { int countUnSetBits=0; while (n) { if ((n & 1) == 0) countUnSetBits++; n=n>>1; } return 1 << countUnSetBits; } int main() { int n = 6; cout<<"The count of integers with equal sum and XOR is "<<countValuesWithEqualSumXOR(n); return 0; }

## Output −

The count of integers with equal sum and XOR is 2

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