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

• Difficulty Level : Hard
• Last Updated : 18 Nov, 2021

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 ``using` `namespace` `std;` `#define MAX 32` `// Function to find A and B``void` `findAB(``int` `N)``{``    ``bitset 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)

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