Related Articles

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

• Difficulty Level : Hard
• Last Updated : 30 Jul, 2019

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

 `// 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
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up