Equal Sum and XOR

Given a positive integer n, count the number of integers i such that: 0 ≤ i ≤ n and n + i = n ^ i, where ^ denotes the bitwise XOR operation.

Examples : 

Input: n = 7
Output: 1
Explanation: The condition holds only for i = 0.
7 + 0 = 7
7 ^ 0 = 7
Therefore, the answer is 1.

Input: n = 12
Output: 4
Explanation: The condition holds for: i = 0, 1, 2, 3
12 + 0 = 12 ^ 0 = 12
12 + 1 = 12 ^ 1 = 13
12 + 2 = 12 ^ 2 = 14
12 + 3 = 12 ^ 3 = 15
Therefore, the answer is 4.

[Naive Approach] Check Every Value - O(n) Time and O(1) Space

This is the simple simulation, we can check each & every value to get the output.

#include <iostream>
using namespace std;

int countValues(int n) {
    int count = 0;

    for (int i = 0; i <= n; i++) {
        if ((n + i) == (n ^ i)) {
            count++;
        }
    }

    return count;
}

int main() {
    cout << countValues(7) << endl;
    cout << countValues(12) << endl;

    return 0;
}

#include <stdio.h>

int main() {

    // code
    return 0;
}

class GFG {
    static int countValues(int n) {
        int count = 0;

        for (int i = 0; i <= n; i++) {
            if ((n + i) == (n ^ i)) {
                count++;
            }
        }

        return count;
    }

    public static void main(String[] args) {
        System.out.println(countValues(7));
        System.out.println(countValues(12));
    }
}

def countValues(n):
    count = 0

    for i in range(n + 1):
        if (n + i) == (n ^ i):
            count += 1

    return count

if __name__ == "__main__":
    print(countValues(7))
    print(countValues(12))

using System;

class GFG {
    static int countValues(int n) {
        int count = 0;

        for (int i = 0; i <= n; i++) {
            if ((n + i) == (n ^ i)) {
                count++;
            }
        }

        return count;
    }

    static void Main() {
        Console.WriteLine(countValues(7));
        Console.WriteLine(countValues(12));
    }
}

function countValues(n) {
    let count = 0;

    for (let i = 0; i <= n; i++) {
        if ((n + i) === (n ^ i)) {
            count++;
        }
    }

    return count;
}

// Driver Code
console.log(countValues(7));
console.log(countValues(12));

1
4

[Expected Approach] Count Unset Bits - O(log n) Time and O(1) Space

If n + i = n ^ i, then there must be no carry while adding n and i. So, i can have set bits only at positions where n has unset bits.

If the binary representation of n has c unset bits, then total valid values are 2^c.

We know that (n+i) = (n^i) + 2*(n&i). So n + i = n ^ i implies n & i = 0. Hence our problem reduces to finding values of i such that n & i = 0. How to find count of such pairs? We can use the count of unset-bits in the binary representation of n. For n & i to be zero, i must unset all set-bits of n. If the kth bit is set at a particular in n, kth bit in i must be 0 always, else kth bit of i can be 0 or 1
Hence, total such combinations are 2^(count of unset bits in n)
For example, consider n = 12 (Binary representation : 1 1 0 0). 
All possible values of i that can unset all bits of n are 0 0 0/1 0/1 where 0/1 implies either 0 or 1. Number of such values of i are 2^2 = 4. 

#include <iostream>
using namespace std;

int countValues(int n) {
    int unsetBits = 0;

    while (n > 0) {
        if ((n & 1) == 0) {
            unsetBits++;
        }

        n >>= 1;
    }

    return 1 << unsetBits;
}

int main() {
    cout << countValues(7) << endl;
    cout << countValues(12) << endl;

    return 0;
}

#include <stdio.h>

int countValues(int n) {
    int unsetBits = 0;

    while (n > 0) {
        if ((n & 1) == 0) {
            unsetBits++;
        }

        n >>= 1;
    }

    return 1 << unsetBits;
}

int main() {
    printf("%d\n", countValues(7));
    printf("%d\n", countValues(12));

    return 0;
}

class GFG {
    static int countValues(int n) {
        int unsetBits = 0;

        while (n > 0) {
            if ((n & 1) == 0) {
                unsetBits++;
            }

            n >>= 1;
        }

        return 1 << unsetBits;
    }

    public static void main(String[] args) {
        System.out.println(countValues(7));
        System.out.println(countValues(12));
    }
}

def countValues(n):
    unsetBits = 0

    while n > 0:
        if (n & 1) == 0:
            unsetBits += 1

        n >>= 1

    return 1 << unsetBits

if __name__ == "__main__":
    print(countValues(7))
    print(countValues(12))

using System;

class GFG {
    static int countValues(int n) {
        int unsetBits = 0;

        while (n > 0) {
            if ((n & 1) == 0) {
                unsetBits++;
            }

            n >>= 1;
        }

        return 1 << unsetBits;
    }

    static void Main() {
        Console.WriteLine(countValues(7));
        Console.WriteLine(countValues(12));
    }
}

function countValues(n) {
    let unsetBits = 0;

    while (n > 0) {
        if ((n & 1) === 0) {
            unsetBits++;
        }

        n >>= 1;
    }

    return 1 << unsetBits;
}

// Driver Code
console.log(countValues(7));
console.log(countValues(12));

1
4