Find the closest pair from two sorted arrays

2.8

Given two sorted arrays and a number x, find the pair whose sum is closest to x and the pair has an element from each array.

We are given two arrays ar1[0…m-1] and ar2[0..n-1] and a number x, we need to find the pair ar1[i] + ar2[j] such that absolute value of (ar1[i] + ar2[j] – x) is minimum.

Example:

Input:  ar1[] = {1, 4, 5, 7};
        ar2[] = {10, 20, 30, 40};
        x = 32      
Output:  1 and 30

Input:  ar1[] = {1, 4, 5, 7};
        ar2[] = {10, 20, 30, 40};
        x = 50      
Output:  7 and 40

We strongly recommend to minimize your browser and try this yourself first.

A Simple Solution is to run two loops. The outer loop considers every element of first array and inner loop checks for the pair in second array. We keep track of minimum difference between ar1[i] + ar2[j] and x.

We can do it in O(n) time using following steps.
1) Merge given two arrays into an auxiliary array of size m+n using merge process of merge sort. While merging keep another boolean array of size m+n to indicate whether the current element in merged array is from ar1[] or ar2[].

2) Consider the merged array and use the linear time algorithm to find the pair with sum closest to x. One extra thing we need to consider only those pairs which have one element from ar1[] and other from ar2[], we use the boolean array for this purpose.

Can we do it in a single pass and O(1) extra space?
The idea is to start from left side of one array and right side of another array, and use the algorithm same as step 2 of above approach. Following is 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 in ar1:  l = 0
       (b) Initialize second  the rightmost index in ar2:  r = n-1
3) Loop while l < m and r >= 0
       (a) If  abs(ar1[l] + ar2[r] - sum) < diff  then 
           update diff and result 
       (b) Else if(ar1[l] + ar2[r] <  sum )  then l++
       (c) Else r--    
4) Print the result. 

Following is C++ implementation of this approach.

C++

// C++ program to find the pair from two sorted arays such
// that the sum of pair is closest to a given number x
#include <iostream>
#include <climits>
#include <cstdlib>
using namespace std;

// ar1[0..m-1] and ar2[0..n-1] are two given sorted arrays
// and x is given number. This function prints the pair  from
// both arrays such that the sum of the pair is closest to x.
void printClosest(int ar1[], int ar2[], int m, int n, int x)
{
    // Initialize the diff between pair sum and x.
    int diff = INT_MAX;

    // res_l and res_r are result indexes from ar1[] and ar2[]
    // respectively
    int res_l, res_r;

    // Start from left side of ar1[] and right side of ar2[]
    int l = 0, r = n-1;
    while (l<m && r>=0)
    {
       // If this pair is closer to x than the previously
       // found closest, then update res_l, res_r and diff
       if (abs(ar1[l] + ar2[r] - x) < diff)
       {
           res_l = l;
           res_r = r;
           diff = abs(ar1[l] + ar2[r] - x);
       }

       // If sum of this pair is more than x, move to smaller
       // side
       if (ar1[l] + ar2[r] > x)
           r--;
       else  // move to the greater side
           l++;
    }

    // Print the result
    cout << "The closest pair is [" << ar1[res_l] << ", "
         << ar2[res_r] << "] \n";
}

// Driver program to test above functions
int main()
{
    int ar1[] = {1, 4, 5, 7};
    int ar2[] = {10, 20, 30, 40};
    int m = sizeof(ar1)/sizeof(ar1[0]);
    int n = sizeof(ar2)/sizeof(ar2[0]);
    int x = 38;
    printClosest(ar1, ar2, m, n, x);
    return 0;
}

Java

// Java program to find closest pair in an array
class ClosestPair
{
    // ar1[0..m-1] and ar2[0..n-1] are two given sorted
    // arrays/ and x is given number. This function prints
    // the pair from both arrays such that the sum of the
    // pair is closest to x.
    void printClosest(int ar1[], int ar2[], int m, int n, int x)
    {
        // Initialize the diff between pair sum and x.
        int diff = Integer.MAX_VALUE;

        // res_l and res_r are result indexes from ar1[] and ar2[]
        // respectively
        int res_l = 0, res_r = 0;

        // Start from left side of ar1[] and right side of ar2[]
        int l = 0, r = n-1;
        while (l<m && r>=0)
        {
           // If this pair is closer to x than the previously
           // found closest, then update res_l, res_r and diff
           if (Math.abs(ar1[l] + ar2[r] - x) < diff)
           {
               res_l = l;
               res_r = r;
               diff = Math.abs(ar1[l] + ar2[r] - x);
           }

           // If sum of this pair is more than x, move to smaller
           // side
           if (ar1[l] + ar2[r] > x)
               r--;
           else  // move to the greater side
               l++;
        }

        // Print the result
        System.out.print("The closest pair is [" + ar1[res_l] +
                         ", " + ar2[res_r] + "]");
    }

    // Driver program to test above functions
    public static void main(String args[])
    {
        ClosestPair ob = new ClosestPair();
        int ar1[] = {1, 4, 5, 7};
        int ar2[] = {10, 20, 30, 40};
        int m = ar1.length;
        int n = ar2.length;
        int x = 38;
        ob.printClosest(ar1, ar2, m, n, x);
    }
}
/*This code is contributed by Rajat Mishra */


Output:
 The closest pair is [7, 30] 

Smallest Difference pair of values between two unsorted Arrays

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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