Equal Sum and XOR - GeeksforGeeks

Equal Sum and XOR - GeeksforGeeks
Given a positive integer n, find count of positive integers i such that 0 <= i <= n and n+i = n^i

Since we know a + b = a ^ b + a & b

We can write, n + i = n ^ i + 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.

int countValues(int n)

{

    // unset_bits keeps track of count of un-set

    // bits in binary representation of n

    int unset_bits=0;

    while (n)

    {

        if ((n & 1) == 0)

            unset_bits++;

        n=n>>1;

    }

    // Return 2 ^ unset_bits

    return 1 << unset_bits;

}

One simple solution is to iterate over all values of i 0<= i <= n and count all satisfying values.

int countValues (int n)

{

    int countV = 0;

    // Traverse all numbers from 0 to n and

    // increment result only when given condition

    // is satisfied.

    for (int i=0; i<=n; i++ )

        if ((n+i) == (n^i) )

            countV++;

    return countV;

}

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