# Non-negative pairs with sum of Bitwise OR and Bitwise AND equal to N

Given an integer N, the task is to find all non-negative pairs (A, B) such that the sum of Bitwise OR and Bitwise AND of A, B is equal to N, i.e., (A | B) + (A & B) = N.

Examples:

Input: N = 5
Output: (0, 5), (1, 4), (2, 3), (3, 2), (4, 1), (5, 0)
Explanation: All possible pairs satisfying the necessary conditions:

1. (0 | 5) + (0 & 5) = 5 + 0 = 5
2. (1 | 4) + (1 & 4) = 5 + 0 = 5
3. (2 | 3) + (2 & 3) = 3 + 2 = 5
4. (3 | 2) + (3 & 2) = 3 + 2 = 5
5. (4 | 1) + (4 & 1) = 5 + 0 = 5
6. (5 | 0) + (5 & 0) = 5 + 0 = 5

Input: N = 7
Output: (0, 7), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (7, 0)
Explanation: All possible pairs satisfying the necessary conditions:

1. (0 | 7) + (0 & 7) = 7 + 0 =7
2. (1 | 6) + (1 & 6) = 7 + 0 =7
3. (2 | 5) + (2 & 5) = 7 + 0 =7
4. (3 | 4) + (3 & 4) = 7 + 0 =7
5. (4 | 3) + (4 & 3) = 7 + 0 =7
6. (5 | 2) + (5 & 2) = 7 + 0 =7
7. (6 | 1) + (6 & 1) = 7 + 0 = 7
8. (7 | 0) + (7 & 0) = 7 + 0 = 7

Naive Approach: The simplest approach is to iterate over the range [0, N] and print those pairs (A, B) that satisfy the condition (A | B) + (A & B) = N

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is based on the observation that all those non-negative pairs whose sum is equal to N satisfy the given condition. Therefore, iterate over the range [0, N] using the variable i and print the pair i and (N – i).

Below is the implementation of the above approach.

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to print all pairs whose` `// sum of Bitwise OR and AND is N` `void` `findPairs(``int` `N)` `{` `    ``// Iterate from i = 0 to N` `    ``for` `(``int` `i = 0; i <= N; i++) {`   `        ``// Print pairs (i, N - i)` `        ``cout << ``"("` `<< i << ``", "` `             ``<< N - i << ``"), "``;` `    ``}` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 5;` `    ``findPairs(N);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.util.*;` `import` `java.lang.*;`   `class` `GFG{` `    `  `// Function to print all pairs whose ` `// sum of Bitwise OR and AND is N` `static` `void` `findPairs(``int` `N)` `{` `    `  `    ``// Iterate from i = 0 to N ` `    ``for``(``int` `i = ``0``; i <= N; i++) ` `    ``{ ` `        `  `        ``// Print pairs (i, N - i)   ` `        ``System.out.print( ``"("` `+ i + ``", "` `+ ` `                          ``(N - i) + ``"), "``); ` `   ``} ` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `N = ``5``;` `    `  `    ``findPairs(N); ` `}` `}`   `// This code is contributed by ajaykr00kj`

## Python3

 `# Python3 program for the above approach `   `# Function to print all pairs whose ` `# sum of Bitwise OR and AND is N ` `def` `findPairs(N):` `    `  `     ``# Iterate from i = 0 to N ` `     ``for` `i ``in` `range``(``0``, N ``+` `1``):` `        `  `        ``# Print pairs (i, N - i) ` `        ``print``(``"("``, i, ``","``, ` `              ``N ``-` `i, ``"), "``, end ``=` `"") `   `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` `    `  `    ``N ``=` `5` `    `  `    ``findPairs(N) `   `# This code is contributed by ajaykr00kj`

## C#

 `// C# program for the above approach ` `using` `System;` ` `  `class` `GFG{` ` `  `// Function to print all pairs whose ` `// sum of Bitwise OR and AND is N` `static` `void` `findPairs(``int` `N)` `{` `    `  `    ``// Iterate from i = 0 to N ` `    ``for``(``int` `i = 0; i <= N; i++) ` `    ``{ ` `        `  `        ``// Print pairs (i, N - i)   ` `        ``Console.Write( ``"("` `+ i + ``", "` `+ ` `                        ``(N - i) + ``"), "``); ` `   ``} ` `}`   `// Driver code` `public` `static` `void` `Main()` `{` `    ``int` `N = 5;` `     `  `    ``findPairs(N); ` `}` `}`   `// This code is contributed by sanjoy_62`

## Javascript

 ``

Output:

`(0, 5), (1, 4), (2, 3), (3, 2), (4, 1), (5, 0),`

Time Complexity: O(N)
Auxiliary Space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!