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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