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 <bits/stdc++.h>
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

<script>
// JavaScript program for the above approach

// Function to print all pairs whose
// sum of Bitwise OR and AND is N
function findPairs(N)
{

    // Iterate from i = 0 to N
    for(let i = 0; i <= N; i++)
    {

        // Print pairs (i, N - i)
        document.write( "(" + i + ", " +
                          (N - i) + "), ");
   }
}

// Driver Code
  let N = 5;
  findPairs(N);

 // This code is contributed by avijitmondal1998.
</script>

Output:

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

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