Closest numbers from a list of unsorted integers

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.

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

Implementation:

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.
?>

Javascript

<script>

// JavaScript 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
    arr.sort();

    // Compare differences of adjacent
    // pairs to find the minimum difference.
    let minDiff = arr[1] - arr[0];
    for (let 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 ( let i = 1; i < n; i++)
     {
        if ((arr[i] - arr[i-1]) == minDiff)
        {
           document.write("(" + arr[i-1] + ", "
                           + arr[i] + ") , " );
        }
     }
  }


// Driver code

        let arr = [5, 3, 2, 4, 1];
    let n = arr.length;
    printMinDiffPairs(arr , n);

</script>

Output: 

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

Time Complexity: O(n log n)
Auxiliary Space: O(1)

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)