Smallest Difference pair of values between two unsorted Arrays
Last Updated :
23 Jul, 2025
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[] = {1, 3, 15, 11, 2}
B[] = {23, 127, 235, 19, 8}
Output : 3
That is, the pair (11, 8) Input : A[] = {10, 5, 40}
B[] = {50, 90, 80}
Output : 10
That is, the pair (40, 50)
Brute Force Approach:
The brute force approach to solve this problem involves comparing each pair of values, one from each array, and calculating their absolute difference. We then keep track of the smallest absolute difference found so far and return it at the end.
Below is the implementation of the above approach:
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) {
int minDiff = INT_MAX; // Initialize with maximum integer value
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int diff = abs(A[i] - B[j]); // Calculate absolute difference
if (diff < minDiff) {
minDiff = diff; // Update smallest difference found so far
}
}
}
return minDiff;
}
// 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;
}
import java.util.*;
public class Main {
// function to calculate smallest result between two arrays
public static int findSmallestDifference(int[] A, int[] B, int m, int n) {
int minDiff = Integer.MAX_VALUE; // Initialize with maximum integer value
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int diff = Math.abs(A[i] - B[j]); // Calculate absolute difference
if (diff < minDiff) {
minDiff = diff; // Update smallest difference found so far
}
}
}
return minDiff;
}
// 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));
}
}
import sys
# function to calculate Small
# result between two arrays
def findSmallestDifference(A, B, m, n):
minDiff = sys.maxsize # Initialize with maximum integer value
for i in range(m):
for j in range(n):
diff = abs(A[i] - B[j]) # Calculate absolute difference
if diff < minDiff:
minDiff = diff # Update smallest difference found so far
return minDiff
# Driver Code
if __name__ == '__main__':
# Input given array A
A = [1, 2, 11, 5]
# Input given array B
B = [4, 12, 19, 23, 127, 235]
# Calculate size of Both arrays
m = len(A)
n = len(B)
# Call function to print smallest result
print(findSmallestDifference(A, B, m, n))
using System;
class Program
{
// Function to calculate smallest result between two arrays
static int FindSmallestDifference(int[] A, int[] B, int m, int n)
{
int minDiff = int.MaxValue; // Initialize with maximum integer value
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
int diff = Math.Abs(A[i] - B[j]); // Calculate absolute difference
if (diff < minDiff)
{
minDiff = diff; // Update smallest difference found so far
}
}
}
return minDiff;
}
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
Console.WriteLine(FindSmallestDifference(A, B, m, n));
}
}
function findSmallestDifference(A, B) {
let minDiff = Infinity;
for (let i = 0; i < A.length; i++) {
for (let j = 0; j < B.length; j++) {
let diff = Math.abs(A[i] - B[j]); // Calculate absolute difference
if (diff < minDiff) {
minDiff = diff; // Update smallest difference found so far
}
}
}
return minDiff;
}
// Input given array A
let A = [1, 2, 11, 5];
// Input given array B
let B = [4, 12, 19, 23, 127, 235];
// Call function to print smallest result
console.log(findSmallestDifference(A, B));
Time Complexity: O(N^2)
Auxiliary Space: O(1)
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: {1, 2, 11, 15}
- B: {4, 12, 19, 23, 127, 235}
- 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 b is 3. Store this as the min.
- How can we (potentially) make this difference smaller? Well, the value at b is bigger than the value at a, so moving b will only make the difference larger. Therefore, we want to move a.
- 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.
- Now a points to 11 and b points to 4. Move b.
- 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 final sma result
return 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 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 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.
# Python 3 Code to find
# Smallest Difference between
# two Arrays
import sys
# function to calculate
# Small result between
# two arrays
def 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 A
A = [1, 2, 11, 5]
# Input given array B
B = [4, 12, 19, 23, 127, 235]
# Calculate size of Both arrays
m = len(A)
n = len(B)
# Call function to
# print smallest result
print(findSmallestDifference(A, B, m, n))
# This code is contributed by
# Smitha Dinesh Semwal
// C# Code to find Smallest
// Difference between two Arrays
using 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.
<script>
// JavaScript Code to find Smallest
// Difference between two Arrays
// function to calculate Small
// result between two arrays
function findSmallestDifference(A, B, m, n)
{
// Sort both arrays using
// sort function
A.sort((a, b) => a - b);
B.sort((a, b) => a - b);
let a = 0, b = 0;
// Initialize result as max value
let result = Number.MAX_SAFE_INTEGER;
// 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
// Input given array A
let A = [ 1, 2, 11, 5 ];
// Input given array B
let B = [ 4, 12, 19, 23, 127, 235 ];
// Calculate size of Both arrays
let m = A.length;
let n = B.length;
// Call function to print
// smallest result
document.write(findSmallestDifference(
A, B, m, n));
// This code is contributed by Surbhi Tyagi.
</script>
<?php
// PHP Code to find Smallest
// Difference between two Arrays
// function to calculate Small
// result between two arrays
function findSmallestDifference($A, $B,
$m, $n)
{
// Sort both arrays
// using sort function
sort($A);
sort($A, $m);
sort($B);
sort($B, $n);
$a = 0; $b = 0;
$INT_MAX = 1;
// Initialize result
// as max value
$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 Code
{
// Input given array A
$A = array(1, 2, 11, 5);
// Input given array B
$B = array(4, 12, 19, 23, 127, 235);
// Calculate size of Both arrays
$m = sizeof($A) / sizeof($A[0]);
$n = sizeof($B) / sizeof($B[0]);
// Call function to print
// smallest result
echo findSmallestDifference($A, $B, $m, $n);
return 0;
}
// This code is contributed by nitin mittal.
?>
Time Complexity: O(m log m + n log n)
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).
Auxiliary Space: O(1)
This article is contributed by Mr. Somesh Awasthi.
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