# Given a sorted array and a number x, find the pair in array whose sum is closest to x

Given a sorted array and a number x, find a pair in an array whose sum is closest to x.

**Examples:**

Input: arr[] = {10, 22, 28, 29, 30, 40}, x = 54 Output: 22 and 30 Input: arr[] = {1, 3, 4, 7, 10}, x = 15 Output: 4 and 10

A simple solution is to consider every pair and keep track of the closest pair (the absolute difference between pair sum and x is minimum). Finally, print the closest pair. The time complexity of this solution is O(n^{2})

An efficient solution can find the pair in O(n) time. The idea is similar to method 1 of this post. The following is a detailed algorithm.

1) Initialize a variable diff as infinite (Diff is used to store the difference between pair and x). We need to find the minimum diff. 2) Initialize two index variables l and r in the given sorted array. (a) Initialize first to the leftmost index: l = 0 (b) Initialize second the rightmost index: r = n-1 3) Loop while l < r. (a) If abs(arr[l] + arr[r] - sum) < diff then update diff and result (b) If(arr[l] + arr[r] < sum ) then l++ (c) Else r--

Following is the implementation of the above algorithm.

## C++14

`// Simple C++ program to find the pair with sum closest to a given no.` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Prints the pair with sum closest to x` `void` `printClosest(` `int` `arr[], ` `int` `n, ` `int` `x)` `{` ` ` `int` `res_l, res_r; ` `// To store indexes of result pair` ` ` `// Initialize left and right indexes and difference between` ` ` `// pair sum and x` ` ` `int` `l = 0, r = n-1, diff = INT_MAX;` ` ` `// While there are elements between l and r` ` ` `while` `(r > l)` ` ` `{` ` ` `// Check if this pair is closer than the closest pair so far` ` ` `if` `(` `abs` `(arr[l] + arr[r] - x) < diff)` ` ` `{` ` ` `res_l = l;` ` ` `res_r = r;` ` ` `diff = ` `abs` `(arr[l] + arr[r] - x);` ` ` `}` ` ` `// If this pair has more sum, move to smaller values.` ` ` `if` `(arr[l] + arr[r] > x)` ` ` `r--;` ` ` `else` `// Move to larger values` ` ` `l++;` ` ` `}` ` ` `cout <<` `" The closest pair is "` `<< arr[res_l] << ` `" and "` `<< arr[res_r];` `}` `// Driver program to test above functions` `int` `main()` `{` ` ` `int` `arr[] = {10, 22, 28, 29, 30, 40}, x = 54;` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]);` ` ` `printClosest(arr, n, x);` ` ` `return` `0;` `}` `// Code By Mayur Patil` |

## Java

`// Java program to find pair with sum closest to x` `import` `java.io.*;` `import` `java.util.*;` `import` `java.lang.Math;` `class` `CloseSum {` ` ` ` ` `// Prints the pair with sum closest to x` ` ` `static` `void` `printClosest(` `int` `arr[], ` `int` `n, ` `int` `x)` ` ` `{` ` ` `int` `res_l=` `0` `, res_r=` `0` `; ` `// To store indexes of result pair` ` ` ` ` `// Initialize left and right indexes and difference between` ` ` `// pair sum and x` ` ` `int` `l = ` `0` `, r = n-` `1` `, diff = Integer.MAX_VALUE;` ` ` ` ` `// While there are elements between l and r` ` ` `while` `(r > l)` ` ` `{` ` ` `// Check if this pair is closer than the closest pair so far` ` ` `if` `(Math.abs(arr[l] + arr[r] - x) < diff)` ` ` `{` ` ` `res_l = l;` ` ` `res_r = r;` ` ` `diff = Math.abs(arr[l] + arr[r] - x);` ` ` `}` ` ` ` ` `// If this pair has more sum, move to smaller values.` ` ` `if` `(arr[l] + arr[r] > x)` ` ` `r--;` ` ` `else` `// Move to larger values` ` ` `l++;` ` ` `}` ` ` ` ` `System.out.println(` `" The closest pair is "` `+arr[res_l]+` `" and "` `+ arr[res_r]);` `}` ` ` ` ` ` ` `// Driver program to test above function` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `arr[] = {` `10` `, ` `22` `, ` `28` `, ` `29` `, ` `30` `, ` `40` `}, x = ` `54` `;` ` ` `int` `n = arr.length;` ` ` `printClosest(arr, n, x); ` ` ` `}` `}` `/*This code is contributed by Devesh Agrawal*/` |

## Python3

`# Python3 program to find the pair` `# with sum` `# closest to a given no.` `# A sufficiently large value greater` `# than any` `# element in the input array` `MAX_VAL ` `=` `1000000000` `#Prints the pair with sum closest to x` `def` `printClosest(arr, n, x):` ` ` ` ` `# To store indexes of result pair` ` ` `res_l, res_r ` `=` `0` `, ` `0` ` ` ` ` `#Initialize left and right indexes` ` ` `# and difference between` ` ` `# pair sum and x` ` ` `l, r, diff ` `=` `0` `, n` `-` `1` `, MAX_VAL` ` ` ` ` `# While there are elements between l and r` ` ` `while` `r > l:` ` ` `# Check if this pair is closer than the` ` ` `# closest pair so far` ` ` `if` `abs` `(arr[l] ` `+` `arr[r] ` `-` `x) < diff:` ` ` `res_l ` `=` `l` ` ` `res_r ` `=` `r` ` ` `diff ` `=` `abs` `(arr[l] ` `+` `arr[r] ` `-` `x)` ` ` ` ` `if` `arr[l] ` `+` `arr[r] > x:` ` ` `# If this pair has more sum, move to` ` ` `# smaller values.` ` ` `r ` `-` `=` `1` ` ` `else` `:` ` ` `# Move to larger values` ` ` `l ` `+` `=` `1` ` ` ` ` `print` `(` `'The closest pair is {} and {}'` ` ` `.` `format` `(arr[res_l], arr[res_r]))` `# Driver code to test above` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `arr ` `=` `[` `10` `, ` `22` `, ` `28` `, ` `29` `, ` `30` `, ` `40` `]` ` ` `n ` `=` `len` `(arr)` ` ` `x` `=` `54` ` ` `printClosest(arr, n, x)` `# This code is contributed by Tuhin Patra` |

## C#

`// C# program to find pair with sum closest to x` `using` `System;` `class` `GFG {` ` ` ` ` `// Prints the pair with sum closest to x` ` ` `static` `void` `printClosest(` `int` `[]arr, ` `int` `n, ` `int` `x)` ` ` `{` ` ` ` ` `// To store indexes of result pair` ` ` `int` `res_l = 0, res_r = 0;` ` ` `// Initialize left and right indexes and` ` ` `// difference between pair sum and x` ` ` `int` `l = 0, r = n-1, diff = ` `int` `.MaxValue;` ` ` `// While there are elements between l and r` ` ` `while` `(r > l)` ` ` `{` ` ` ` ` `// Check if this pair is closer than the` ` ` `// closest pair so far` ` ` `if` `(Math.Abs(arr[l] + arr[r] - x) < diff)` ` ` `{` ` ` `res_l = l;` ` ` `res_r = r;` ` ` `diff = Math.Abs(arr[l] + arr[r] - x);` ` ` `}` ` ` `// If this pair has more sum, move to` ` ` `// smaller values.` ` ` `if` `(arr[l] + arr[r] > x)` ` ` `r--;` ` ` `else` `// Move to larger values` ` ` `l++;` ` ` `}` ` ` ` ` `Console.Write(` `" The closest pair is "` `+` ` ` `arr[res_l] + ` `" and "` `+ arr[res_r]);` ` ` `}` ` ` ` ` `// Driver program to test above function` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `[]arr = {10, 22, 28, 29, 30, 40};` ` ` `int` `x = 54;` ` ` `int` `n = arr.Length;` ` ` ` ` `printClosest(arr, n, x); ` ` ` `}` `}` `// This code is contributed by nitin mittal.` |

## PHP

`<?php` `// Simple PHP program to find the` `// pair with sum closest to a` `// given no.` `// Prints the pair with` `// sum closest to x` `function` `printClosest(` `$arr` `, ` `$n` `, ` `$x` `)` `{` ` ` ` ` `// To store indexes` ` ` `// of result pair` ` ` `$res_l` `;` ` ` `$res_r` `;` ` ` `// Initialize left and right` ` ` `// indexes and difference between` ` ` `// pair sum and x` ` ` `$l` `= 0;` ` ` `$r` `= ` `$n` `- 1;` ` ` `$diff` `= PHP_INT_MAX;` ` ` `// While there are elements` ` ` `// between l and r` ` ` `while` `(` `$r` `> ` `$l` `)` ` ` `{` ` ` ` ` `// Check if this pair is closer` ` ` `// than the closest pair so far` ` ` `if` `(` `abs` `(` `$arr` `[` `$l` `] + ` `$arr` `[` `$r` `] - ` `$x` `) <` ` ` `$diff` `)` ` ` `{` ` ` `$res_l` `= ` `$l` `;` ` ` `$res_r` `= ` `$r` `;` ` ` `$diff` `= ` `abs` `(` `$arr` `[` `$l` `] + ` `$arr` `[` `$r` `] - ` `$x` `);` ` ` `}` ` ` ` ` `// If this pair has more sum,` ` ` `// move to smaller values.` ` ` `if` `(` `$arr` `[` `$l` `] + ` `$arr` `[` `$r` `] > ` `$x` `)` ` ` `$r` `--;` ` ` ` ` `// Move to larger values` ` ` `else` ` ` `$l` `++;` ` ` `}` ` ` `echo` `" The closest pair is "` ` ` `, ` `$arr` `[` `$res_l` `] ,` `" and "` ` ` `, ` `$arr` `[` `$res_r` `];` `}` ` ` `// Driver Code` ` ` `$arr` `= ` `array` `(10, 22, 28, 29, 30, 40);` ` ` `$x` `= 54;` ` ` `$n` `= ` `count` `(` `$arr` `);` ` ` `printClosest(` `$arr` `, ` `$n` `, ` `$x` `);` ` ` `// This code is contributed by anuj_67.` `?>` |

## Javascript

`<script>` `// JavaScript program to find pair` `// with sum closest to x` ` ` ` ` `// Prints the pair with sum closest to x` ` ` `function` `printClosest(arr,n,x)` ` ` `{` ` ` `// To store indexes of result pair` ` ` `let res_l=0, res_r=0;` ` ` `// Initialize left and right indexes` ` ` `// and difference between` ` ` `// pair sum and x` ` ` `let l = 0, r = n-1, diff = Number.MAX_VALUE;` ` ` `// While there are elements` ` ` `// between l and r` ` ` `while` `(r > l)` ` ` `{` ` ` `// Check if this pair is closer` ` ` `// than the closest pair so far` ` ` `if` `(Math.abs(arr[l] +` ` ` `arr[r] - x) < diff)` ` ` `{` ` ` `res_l = l;` ` ` `res_r = r;` ` ` `diff = Math.abs(arr[l] + arr[r] - x);` ` ` `}` ` ` `// If this pair has more sum,` ` ` `// move to smaller values.` ` ` `if` `(arr[l] + arr[r] > x)` ` ` `r--;` ` ` `else` `// Move to larger values` ` ` `l++;` ` ` `}` ` ` `document.write(` ` ` `" The closest pair is "` `+arr[res_l]+` `" and "` `+ arr[res_r]` ` ` `);` `}` ` ` ` ` ` ` `// Driver program to test above function` ` ` ` ` `let arr = [10, 22, 28, 29, 30, 40], x = 54;` ` ` `let n = arr.length;` ` ` `printClosest(arr, n, x); ` ` ` `// This code is contributed by sravan kumar` `</script>` |

**Output**

The closest pair is 22 and 30

**Time Complexity: O(n)**, where n is the length of an Array.**Auxiliary Space:** O(1)