Find minimum difference between any two elements
Given an unsorted array, find the minimum difference between any pair in given array.
Examples :
Input : {1, 5, 3, 19, 18, 25};
Output : 1
Minimum difference is between 18 and 19
Input : {30, 5, 20, 9};
Output : 4
Minimum difference is between 5 and 9
Input : {1, 19, -4, 31, 38, 25, 100};
Output : 5
Minimum difference is between 1 and -4
Method 1 (Simple: O(n2)
A simple solution is to use two loops.
C++
// C++ implementation of simple method to find// minimum difference between any pair#include <bits/stdc++.h>using namespace std;// Returns minimum difference between any pairint findMinDiff(int arr[], int n){// Initialize difference as infiniteint diff = INT_MAX;// Find the min diff by comparing difference// of all possible pairs in given arrayfor (int i=0; i<n-1; i++) for (int j=i+1; j<n; j++) if (abs(arr[i] - arr[j]) < diff) diff = abs(arr[i] - arr[j]);// Return min diffreturn diff;}// Driver codeint main(){int arr[] = {1, 5, 3, 19, 18, 25};int n = sizeof(arr)/sizeof(arr[0]);cout << "Minimum difference is " << findMinDiff(arr, n);return 0;} |
Java
// Java implementation of simple method to find// minimum difference between any pairclass GFG{ // Returns minimum difference between any pair static int findMinDiff(int[] arr, int n) { // Initialize difference as infinite int diff = Integer.MAX_VALUE; // Find the min diff by comparing difference // of all possible pairs in given array for (int i=0; i<n-1; i++) for (int j=i+1; j<n; j++) if (Math.abs((arr[i] - arr[j]) )< diff) diff = Math.abs((arr[i] - arr[j])); // Return min diff return diff; } // Driver method to test the above function public static void main(String[] args) { int arr[] = new int[]{1, 5, 3, 19, 18, 25}; System.out.println("Minimum difference is "+ findMinDiff(arr, arr.length)); }} |
Python
# Python implementation of simple method to find# minimum difference between any pair# Returns minimum difference between any pairdef findMinDiff(arr, n): # Initialize difference as infinite diff = 10**20 # Find the min diff by comparing difference # of all possible pairs in given array for i in range(n-1): for j in range(i+1,n): if abs(arr[i]-arr[j]) < diff: diff = abs(arr[i] - arr[j]) # Return min diff return diff# Driver codearr = [1, 5, 3, 19, 18, 25]n = len(arr)print("Minimum difference is " + str(findMinDiff(arr, n)))# This code is contributed by Pratik Chhajer |
C#
// C# implementation of simple method to find// minimum difference between any pairusing System;class GFG { // Returns minimum difference between any pair static int findMinDiff(int []arr, int n) { // Initialize difference as infinite int diff = int.MaxValue; // Find the min diff by comparing difference // of all possible pairs in given array for (int i = 0; i < n-1; i++) for (int j = i+1; j < n; j++) if (Math.Abs((arr[i] - arr[j]) ) < diff) diff = Math.Abs((arr[i] - arr[j])); // Return min diff return diff; } // Driver method to test the above function public static void Main() { int []arr = new int[]{1, 5, 3, 19, 18, 25}; Console.Write("Minimum difference is " + findMinDiff(arr, arr.Length)); }}// This code is contributed by nitin mittal. |
PHP
<?php// PHP implementation of simple// method to find minimum// difference between any pair// Returns minimum difference// between any pairfunction findMinDiff($arr, $n){// Initialize difference// as infinite$diff = PHP_INT_MAX;// Find the min diff by comparing// difference of all possible// pairs in given arrayfor ($i = 0; $i < $n - 1; $i++) for ($j = $i + 1; $j < $n; $j++) if (abs($arr[$i] - $arr[$j]) < $diff) $diff = abs($arr[$i] - $arr[$j]);// Return min diffreturn $diff;}// Driver code$arr = array(1, 5, 3, 19, 18, 25);$n = sizeof($arr);echo "Minimum difference is " , findMinDiff($arr, $n);// This code is contributed by ajit?> |
Javascript
<script>// JavaScript program implementation of simple method to find// minimum difference between any pair // Returns minimum difference between any pair function findMinDiff( arr, n) { // Initialize difference as infinite let diff = Number.MAX_VALUE; // Find the min diff by comparing difference // of all possible pairs in given array for (let i=0; i<n-1; i++) for (let j=i+1; j<n; j++) if (Math.abs((arr[i] - arr[j]) )< diff) diff = Math.abs((arr[i] - arr[j])); // Return min diff return diff; }// Driver Code let arr = [1, 5, 3, 19, 18, 25]; document.write("Minimum difference is "+ findMinDiff(arr, arr.length));</script> |
Output :
Minimum difference is 1
Method 2 (Efficient: O(n Log n)
The idea is to use sorting. Below are steps.
1) Sort array in ascending order. This step takes O(n Log n) time.
2) Initialize difference as infinite. This step takes O(1) time.
3) Compare all adjacent pairs in sorted array and keep track of minimum difference. This step takes O(n) time.
Below is implementation of above idea.
C++
// C++ program to find minimum difference between// any pair in an unsorted array#include <bits/stdc++.h>using namespace std;// Returns minimum difference between any pairint findMinDiff(int arr[], int n){ // Sort array in non-decreasing order sort(arr, arr+n); // Initialize difference as infinite int diff = INT_MAX; // Find the min diff by comparing adjacent // pairs in sorted array for (int i=0; i<n-1; i++) if (arr[i+1] - arr[i] < diff) diff = arr[i+1] - arr[i]; // Return min diff return diff;}// Driver codeint main(){ int arr[] = {1, 5, 3, 19, 18, 25}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Minimum difference is " << findMinDiff(arr, n); return 0;} |
Java
// Java program to find minimum difference between// any pair in an unsorted arrayimport java.util.Arrays;class GFG{ // Returns minimum difference between any pair static int findMinDiff(int[] arr, int n) { // Sort array in non-decreasing order Arrays.sort(arr); // Initialize difference as infinite int diff = Integer.MAX_VALUE; // Find the min diff by comparing adjacent // pairs in sorted array for (int i=0; i<n-1; i++) if (arr[i+1] - arr[i] < diff) diff = arr[i+1] - arr[i]; // Return min diff return diff; } // Driver method to test the above function public static void main(String[] args) { int arr[] = new int[]{1, 5, 3, 19, 18, 25}; System.out.println("Minimum difference is "+ findMinDiff(arr, arr.length)); }} |
Python
# Python program to find minimum difference between# any pair in an unsorted array# Returns minimum difference between any pairdef findMinDiff(arr, n): # Sort array in non-decreasing order arr = sorted(arr) # Initialize difference as infinite diff = 10**20 # Find the min diff by comparing adjacent # pairs in sorted array for i in range(n-1): if arr[i+1] - arr[i] < diff: diff = arr[i+1] - arr[i] # Return min diff return diff# Driver codearr = [1, 5, 3, 19, 18, 25]n = len(arr)print("Minimum difference is " + str(findMinDiff(arr, n)))# This code is contributed by Pratik Chhajer |
C#
// C# program to find minimum// difference between any pair// in an unsorted arrayusing System;class GFG{ // Returns minimum difference // between any pair static int findMinDiff(int[] arr, int n) { // Sort array in // non-decreasing order Array.Sort(arr); // Initialize difference // as infinite int diff = int.MaxValue; // Find the min diff by // comparing adjacent pairs // in sorted array for (int i = 0; i < n - 1; i++) if (arr[i + 1] - arr[i] < diff) diff = arr[i + 1] - arr[i]; // Return min diff return diff; } // Driver Code public static void Main() { int []arr = new int[]{1, 5, 3, 19, 18, 25}; Console.WriteLine("Minimum difference is " + findMinDiff(arr, arr.Length)); }}//This code is contributed by anuj_67. |
PHP
<?php// PHP program to find minimum// difference between any pair// in an unsorted array// Returns minimum difference// between any pairfunction findMinDiff($arr, $n){ // Sort array in// non-decreasing ordersort($arr);// Initialize difference// as infinite$diff = PHP_INT_MAX;// Find the min diff by// comparing adjacent// pairs in sorted arrayfor ($i = 0; $i < $n - 1; $i++) if ($arr[$i + 1] - $arr[$i] < $diff) $diff = $arr[$i + 1] - $arr[$i];// Return min diffreturn $diff;}// Driver code$arr = array(1, 5, 3, 19, 18, 25);$n = sizeof($arr);echo "Minimum difference is " , findMinDiff($arr, $n);// This code is contributed ajit?> |
Javascript
<script> // Javascript program to find minimum // difference between any pair // in an unsorted array // Returns minimum difference // between any pair function findMinDiff(arr, n) { // Sort array in // non-decreasing order arr.sort(function(a, b) {return a - b}); // Initialize difference // as infinite let diff = Number.MAX_VALUE; // Find the min diff by // comparing adjacent pairs // in sorted array for (let i = 0; i < n - 1; i++) if (arr[i + 1] - arr[i] < diff) diff = arr[i + 1] - arr[i]; // Return min diff return diff; } let arr = [1, 5, 3, 19, 18, 25]; document.write("Minimum difference is " + findMinDiff(arr, arr.length)); </script> |
Output :
Minimum difference is 1
This article is contributed by Harshit Agrawal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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