Given a list of distinct unsorted integers, find the pair of elements that have the smallest absolute difference between them? If there are multiple pairs, find them all.

Examples:

Input : arr[] = {10, 50, 12, 100} Output : (10, 12) The closest elements are 10 and 12 Input : arr[] = {5, 4, 3, 2} Output : (2, 3), (3, 4), (4, 5)

This problem is mainly an extension of Find minimum difference between any two elements.

- Sort the given array.
- Find minimum difference of all pairs in linear time in sorted array.
- Traverse sorted array one more time to print all pairs with minimum difference obtained in step 2.

## C++

// CPP program to find minimum difference // an unsorted array. #include<bits/stdc++.h> using namespace std; // Returns minimum difference between any // two pair in arr[0..n-1] void printMinDiffPairs(int arr[], int n) { if (n <= 1) return; // Sort array elements sort(arr, arr+n); // Compare differences of adjacent // pairs to find the minimum difference. int minDiff = arr[1] - arr[0]; for (int i = 2 ; i < n ; i++) minDiff = min(minDiff, arr[i] - arr[i-1]); // Traverse array again and print all pairs // with difference as minDiff. for (int i = 1; i < n; i++) if ((arr[i] - arr[i-1]) == minDiff) cout << "(" << arr[i-1] << ", " << arr[i] << "), "; } // Driver code int main() { int arr[] = {5, 3, 2, 4, 1}; int n = sizeof(arr) / sizeof(arr[0]); printMinDiffPairs(arr, n); return 0; }

## Java

// Java program to find minimum // difference an unsorted array. import java.util.*; class GFG { // Returns minimum difference between // any two pair in arr[0..n-1] static void printMinDiffPairs(int arr[], int n) { if (n <= 1) return; // Sort array elements Arrays.sort(arr); // Compare differences of adjacent // pairs to find the minimum difference. int minDiff = arr[1] - arr[0]; for (int i = 2; i < n; i++) minDiff = Math.min(minDiff, arr[i] - arr[i-1]); // Traverse array again and print all pairs // with difference as minDiff. for ( int i = 1; i < n; i++) { if ((arr[i] - arr[i-1]) == minDiff) { System.out.print("(" + arr[i-1] + ", " + arr[i] + ")," ); } } } // Driver code public static void main (String[] args) { int arr[] = {5, 3, 2, 4, 1}; int n = arr.length; printMinDiffPairs(arr, n); } } // This code is contributed by Ansu Kumari

## Python3

# Python3 program to find minimum # difference in an unsorted array. # Returns minimum difference between # any two pair in arr[0..n-1] def printMinDiffPairs(arr , n): if n <= 1: return # Sort array elements arr.sort() # Compare differences of adjacent # pairs to find the minimum difference. minDiff = arr[1] - arr[0] for i in range(2 , n): minDiff = min(minDiff, arr[i] - arr[i-1]) # Traverse array again and print all # pairs with difference as minDiff. for i in range(1 , n): if (arr[i] - arr[i-1]) == minDiff: print( "(" + str(arr[i-1]) + ", " + str(arr[i]) + "), ", end = '') # Driver code arr = [5, 3, 2, 4, 1] n = len(arr) printMinDiffPairs(arr , n) # This code is contributed by Ansu Kumari

## C#

// C# program to find minimum // difference an unsorted array. using System; class GFG { // Returns minimum difference between // any two pair in arr[0..n-1] static void printMinDiffPairs(int []arr, int n) { if (n <= 1) return; // Sort array elements Array.Sort(arr); // Compare differences of adjacent // pairs to find the minimum difference. int minDiff = arr[1] - arr[0]; for (int i = 2; i < n; i++) minDiff = Math.Min(minDiff, arr[i] - arr[i-1]); // Traverse array again and print all pairs // with difference as minDiff. for ( int i = 1; i < n; i++) { if ((arr[i] - arr[i-1]) == minDiff) { Console.Write(" (" + arr[i-1] + ", " + arr[i] + "), " ); } } } // Driver code public static void Main () { int []arr = {5, 3, 2, 4, 1}; int n = arr.Length; printMinDiffPairs(arr, n); } } // This code is contributed by vt_m

## PHP

<?php //PHP program to find minimum difference // an unsorted array. // Returns minimum difference between any // two pair in arr[0..n-1] function printMinDiffPairs($arr, $n) { if ($n <= 1) return; // Sort array elements sort($arr); // Compare differences of adjacent // pairs to find the minimum // difference. $minDiff = $arr[1] - $arr[0]; for ($i = 2 ; $i < $n ; $i++) $minDiff = min($minDiff, $arr[$i] - $arr[$i-1]); // Traverse array again and print all // pairs with difference as minDiff. for ($i = 1; $i < $n; $i++) if (($arr[$i] - $arr[$i-1]) == $minDiff) echo "(" , $arr[$i-1] , ", ", $arr[$i] , "), "; } // Driver code $arr = array(5, 3, 2, 4, 1); $n = sizeof($arr); printMinDiffPairs($arr, $n); // This code is contributed by ajit. ?>

Output:

(1, 2), (2, 3), (3, 4), (4, 5),

**Does above program handle duplicates?**

The cases like {x, x, x} are not handled by above program. For this case, the expected output (x, x), (x, x), (x, x), but above program prints (x, x), (x, x)

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