Smallest Difference pair of values between two unsorted Arrays

2.5

Given two arrays of integers, compute the pair of values (one value in each array) with the smallest (non-negative) difference. Return the difference.

Examples:

Input : A[] = {l, 3, 15, 11, 2}
        B[] = {23, 127, 235, 19, 8} 
Output : 3  
That is, the pair (11, 8) 

Input : A[] = {l0, 5, 40}
        B[] = {50, 90, 80} 
Output : 10
That is, the pair (40, 50)

A simple solution is to Brute Force using two loops with Time Complexity O(n2).

A better solution is to sort the arrays. Once the arrays are sorted, we can find the minimum difference by iterating through the arrays using the approach discussed in below post.

Find the closest pair from two sorted arrays

Consider the following two arrays:
A: {l, 2, 11, 15}
B: {4, 12, 19, 23, 127, 235}

1. Suppose a pointer a points to the beginning of A and a pointer b points to the beginning of B. The current difference between a and bis 3. Store this as the min.

2. How can we (potentially) make this difference smaller? Well, the value at bis bigger than the value at a, so moving b will only make the difference larger. Therefore, we want to move a.

3. Now a points to 2 and b (still) points to 4. This difference is 2, so we should update min. Move a, since it is smaller.

4. Now a points to 11 and b points to 4. Move b.

5. Now a points to 11 and b points to 12. Update min to 1. Move b. And so on.

Below is the implementation of the idea.

C++

// C++ Code to find Smallest Difference between
// two Arrays
#include <bits/stdc++.h>
using namespace std;

// function to calculate Small result between
// two arrays
int findSmallestDifference(int A[], int B[],
                               int m, int n)
{
    // Sort both arrays using sort function
    sort(A, A+m);
    sort(B, B+n);

    int a = 0, b = 0;

    // Initialize result as max value
    int result = INT_MAX;

    // Scan Both Arrays upto sizeof of the Arrays
    while (a<m && b<n)
    {
        if (abs(A[a]-B[b]) < result)
            result = abs(A[a]-B[b]);

        // Move Smaller Value
        if (A[a] < B[b])
            a++;

        else
            b++;
    }

    return result; // return  final sma  result
}

// Driver Code
int main()
{
    // Input given array A
    int A[] = {1, 2, 11, 5};

    // Input given array B
    int B[] = {4, 12, 19, 23, 127, 235};


    // Calculate size of Both arrays
    int m = sizeof(A)/sizeof(A[0]);
    int n = sizeof(B)/sizeof(B[0]);

    // Call function to print smallest result
    cout << findSmallestDifference(A, B, m, n);

    return 0;
}

Java

// Java Code to find Smallest Difference between
// two Arrays
import java.util.*;

class GFG {
    
    // function to calculate Small result between
    // two arrays
    static int findSmallestDifference(int A[], int B[],
                                      int m, int n)
    {
        // Sort both arrays using sort function
        Arrays.sort(A);
        Arrays.sort(B);
     
        int a = 0, b = 0;
     
        // Initialize result as max value
        int result = Integer.MAX_VALUE;
     
        // Scan Both Arrays upto sizeof of the 
        // Arrays
        while (a < m && b < n)
        {
            if (Math.abs(A[a] - B[b]) < result)
                result = Math.abs(A[a] - B[b]);
     
            // Move Smaller Value
            if (A[a] < B[b])
                a++;
     
            else
                b++;
        }
        
        // return  final sma  result
        return result; 
    }
    
    /* Driver program to test above function */
    public static void main(String[] args) 
    {
        // Input given array A
        int A[] = {1, 2, 11, 5};
     
        // Input given array B
        int B[] = {4, 12, 19, 23, 127, 235};
     
     
        // Calculate size of Both arrays
        int m = A.length;
        int n = B.length;
     
        // Call function to print smallest result
        System.out.println(findSmallestDifference
                             (A, B, m, n));
       
    }
}
// This code is contributed by Arnav Kr. Mandal.


Output:

1

This algorithm takes O(m log m + n log n) time to sort and O(m + n) time to find the minimum difference. Therefore, the overall runtime is O(m log m + n log n).

This article is contributed by Mr. Somesh Awasthi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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