Find a pair with sum N having minimum absolute difference

Given an integer N, the task is to find a distinct pair of X and Y such that X + Y = N and abs(X - Y) is minimum.

Examples:

Input: N = 11
Output: 5 6 
Explanation:
X = 5 and Y = 6 satisfy the given equation. 
Therefore, the minimum absolute value of abs(X - Y) = 1.

Input: N = 12 
Output: 5 7

Naive Approach: The simplest approach to solve this problem is to generate all possible values of X and Y with a sum equal to N and print the value of X and Y which gives the minimum absolute value of abs(X - Y).

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

Efficient Approach: The above approach can be optimized based on the following observations:

If N % 2 == 1, then pairs (N / 2) and (N / 2 + 1) have minimum absolute difference. 
Otherwise, pairs (N / 2 - 1) and (N / 2 + 1) will have the minimum absolute difference.

Follow the steps below to solve the problem:

• Check if N is odd or not. If found to be true, then print the floor value of (N / 2) and (N / 2 + 1) as the required answer.
• Otherwise, print the value of (N / 2 - 1) and (N / 2 + 1).

Below is the implementation of the above approach:

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the value of X and Y
// having minimum value of abs(X - Y)
void findXandYwithminABSX_Y(int N)
{
    // If N is an odd number
    if (N % 2 == 1) {
        cout << (N / 2) << " " << (N / 2 + 1);
    }

    // If N is an even number
    else {
        cout << (N / 2 - 1) << " " << (N / 2 + 1);
    }
}

// Driver Code
int main()
{
    int N = 12;
    findXandYwithminABSX_Y(N);
}

// Java program to implement
// the above approach
import java.util.*;
class GFG {

    // Function to find the value
    // of X and Y having minimum
    // value of Math.abs(X - Y)
    static void findXandYwithminABSX_Y(int N)
    {
        // If N is an odd number
        if (N % 2 == 1) {
            System.out.print((N / 2) + " " + (N / 2 + 1));
        }

        // If N is an even number
        else {
            System.out.print((N / 2 - 1) + " "
                             + (N / 2 + 1));
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 12;
        findXandYwithminABSX_Y(N);
    }
}

// This code is contributed by gauravrajput1

# Python3 program to implement
# the above approach

# Function to find the value of X and Y
# having minimum value of abs(X - Y)


def findXandYwithminABSX_Y(N):

    # If N is an odd number
    if (N % 2 == 1):
        print((N // 2), (N // 2 + 1))

    # If N is an even number
    else:
        print((N // 2 - 1), (N // 2 + 1))


# Driver Code
if __name__ == '__main__':

    N = 12

    findXandYwithminABSX_Y(N)

# This code is contributed by mohit kumar 29

// C# program to implement
// the above approach
using System;
class GFG {

    // Function to find the value
    // of X and Y having minimum
    // value of Math.abs(X - Y)
    static void findXandYwithminABSX_Y(int N)
    {
        // If N is an odd number
        if (N % 2 == 1) {
            Console.Write((N / 2) + " " + (N / 2 + 1));
        }

        // If N is an even number
        else {
            Console.Write((N / 2 - 1) + " " + (N / 2 + 1));
        }
    }

    // Driver Code
    public static void Main()
    {
        int N = 12;
        findXandYwithminABSX_Y(N);
    }
}

// This code is contributed by bgangwar59

<?php
  
function findXandYwithminABSX_Y($N){
  
    // If N is an odd number
    if ($N % 2 == 1)
  {
    return ($N / 2) . " " . ($N / 2 + 1);
  }
 
  // If N is an even number
  else
  {
    return ($N /2 -1) . " " . ($N / 2 + 1);
  }
  
  
}

  
// Driver code   
$N = 12;
echo(findXandYwithminABSX_Y($N));

  
?>

<script>

// JavaScript program to implement the above approach

// Function to find the value of X and Y
// having minimum value of abs(X - Y)
function findXandYwithminABSX_Y(N)
{

    // If N is an odd number
    if (N % 2 == 1) 
    {
        document.write((N / 2) + " " + (N / 2 + 1));
    }

    // If N is an even number
    else 
    {
        document.write((N / 2 - 1) + " " + (N / 2 + 1));
    }
}


// Driver Code 

    let N = 12;
    findXandYwithminABSX_Y(N);
    
    // This code is contributed by susmitakundugoaldanga.
</script>

5 7

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

using  Brute Force in python:

Approach:

• Initialize a variable min_diff to infinity.
• Use two nested loops to generate all possible pairs of numbers (i, j) such that 1 <= i < j <= N.
• For each pair (i, j), check if i + j == N.
• If i + j == N, calculate the absolute difference between i and j using abs(i - j).
• If the absolute difference is less than the current minimum difference min_diff, update min_diff to the new absolute difference and store the pair (i, j) as the new answer.
• After checking all possible pairs, return the answer as a tuple (x, y) where x and y are the pair of numbers that add up to N with the minimum absolute difference.

def min_abs_diff_pair_brute_force(N):
    min_diff = float('inf')
    for i in range(1, N):
        for j in range(i+1, N+1):
            if abs(i-j) < min_diff and i+j == N:
                min_diff = abs(i-j)
                x, y = i, j
    return x, y

# Test the function with the given inputs
print(min_abs_diff_pair_brute_force(11))  # Output: (5, 6)
print(min_abs_diff_pair_brute_force(12))  # Output: (5, 7)

(5, 6)
(5, 7)

time complexity of O(N^2)
space complexity of O(1),