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 = 11Output:5 6Explanation:

X = 5 and Y = 6 satisfy the given equation.

Therefore, the minimum absolute value of abs(X – Y) = 1.

Input:N = 12Output: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(N^{2}) **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++

`// 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

`// 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

`# 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#

`// 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

`<?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` `));` ` ` `?>` |

## Javascript

`<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>` |

**Output:**

5 7

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

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