Find two integers A and B such that A ^ N = A + N and B ^ N = B + N

Given a non-negative integer N, the task is to find two integers A (greatest integer smaller than N) and (smallest integer greater than N) such that A + N = A ^ N and B + N = B ^ N
Examples:

Input: N = 5 
Output: A = 2 and B = 8 
2 + 8 = 2 ^ 8 = 10
Input: N = 10 
Output: A = 5 and B = 16 
5 + 16 = 5 ^ 16 = 21 

Approach: Lets find A and B independently. To solve this problem we have to use the property, x + y = x^y + 2 * (x & y)
Since the problem states that xor sum is equal to the given sum which implies that their AND must be 0. 

• Finding A: N can be represented as a series of bits of 0 and 1. To find A we will first have to find the most significant bit of N which is set. Since A & N = 0, The places where N has set bit, for that places we will make bits of A as unset and for the places where N has unset bit, we will make that bit set for A as we want to maximize A. This we will do for all the bits from most significant to the least significant. Hence we will get our A.
• Finding B: Finding B is easy. Let i be the position of the leftmost set bit in 1. We want B to be greater than N, also we want B & N =0. Hence using these two facts B will be always (1<< (i+1)).

Below is the implementation of the above approach:

CPP

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

#define MAX 32

// Function to find A and B
void findAB(int N)
{
    bitset<MAX> arr(N), brr(N);

    // To store the leftmost set bit in N
    int leftsetN = -1;
    for (int i = MAX - 1; i >= 0; --i) {
        if (arr[i] == 1) {
            leftsetN = i;
            break;
        }
    }

    // To store the value of A
    int A = 0;
    for (int i = leftsetN; i >= 0; --i) {

        // If the bit is unset in N
        // then  we will set it in A
        if (arr[i] == 0) {
            A |= (1 << i);
        }
    }

    // To store the value of B
    int B = 0;

    // B will be (1 << (leftsetN + 1))
    B = 1 << (leftsetN + 1);

    // Print the values of A and B
    cout << "A = " << A << " and B = " << B;
}

// Driver code
int main()
{
    int N = 5;

    findAB(N);

    return 0;
}

Output:

A = 2 and B = 8

Time Complexity: O(MAX)

Auxiliary Space: O(N)