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

Difficulty Level : Easy
Last Updated : 06 May, 2021

Given a sorted array and a number x, find a pair in 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 closest pair (absolute difference between pair sum and x is minimum). Finally, print the closest pair. Time complexity of this solution is O(n²)

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++

// 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;
}

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 cloest 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 cloest 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 cloest 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

This article is contributed by Harsh. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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