Smallest Difference pair of values between two unsorted Arrays

• Difficulty Level : Easy
• Last Updated : 18 May, 2021

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 using namespace std; // function to calculate Small// result between two arraysint 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 final sma result    return result;} // Driver Codeint 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);    int n = sizeof(B) / sizeof(B);     // Call function to print    // smallest result    cout << findSmallestDifference(A, B, m, n);     return 0;}

Java

 // Java Code to find Smallest// Difference between two Arraysimport 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 Code    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.

Python3

 # Python 3 Code to find# Smallest Difference between# two Arraysimport sys # function to calculate# Small result between# two arraysdef findSmallestDifference(A, B, m, n):     # Sort both arrays    # using sort function    A.sort()    B.sort()     a = 0    b = 0     # Initialize result as max value    result = sys.maxsize     # Scan Both Arrays upto    # sizeof of the Arrays    while (a < m and b < n):             if (abs(A[a] - B[b]) < result):            result = abs(A[a] - B[b])         # Move Smaller Value        if (A[a] < B[b]):            a += 1         else:            b += 1    # return final sma result    return result # Driver Code # Input given array AA = [1, 2, 11, 5] # Input given array BB = [4, 12, 19, 23, 127, 235] # Calculate size of Both arraysm = len(A)n = len(B) # Call function to# print smallest resultprint(findSmallestDifference(A, B, m, n)) # This code is contributed by# Smitha Dinesh Semwal

C#

 // C# Code to find Smallest// Difference between two Arraysusing System; 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        Array.Sort(A);        Array.Sort(B);             int a = 0, b = 0;             // Initialize result as max value        int result = int.MaxValue;             // 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 Code    public static void 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 = A.Length;        int n = B.Length;             // Call function to        // print smallest result        Console.Write(findSmallestDifference                              (A, B, m, n));             }} // This code is contributed// by nitin mittal.



Javascript



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).

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