Find minimum difference between any two elements | Set 2
Given an unsorted array arr[] of size n, the task is to find the minimum difference between any pair in the given array.
Input: arr[] = {1, 2, 3, 4}
Output: 1
The possible absolute differences are:
{1, 2, 3, 1, 2, 1}Input: arr[] = {10, 2, 5, 4}
Output: 1
Approach:
- Traverse through the array and create a hash array to store the frequency of the array elements.
- Now, traverse through the hash array and calculate the distance between two nearest elements.
- The frequency was calculated to check whether some element has frequency > 1 which means the absolute distance will be 0 i.e. |arr[i] – arr[i]|.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MAX 100001 // Function to return the minimum// absolute difference between any// two elements of the arrayint getMinDiff(int arr[], int n){ // To store the frequency of each element int freq[MAX] = { 0 }; for (int i = 0; i < n; i++) { // Update the frequency of current element freq[arr[i]]++; // If current element appears more than once // then the minimum absolute difference // will be 0 i.e. |arr[i] - arr[i]| if (freq[arr[i]] > 1) return 0; } int mn = INT_MAX; // Checking the distance between the nearest // two elements in the frequency array for (int i = 0; i < MAX; i++) { if (freq[i] > 0) { i++; int cnt = 1; while ((freq[i] == 0) && (i != MAX - 1)) { cnt++; i++; } mn = min(cnt, mn); i--; } } // Return the minimum absolute difference return mn;} // Driver codeint main(){ int arr[] = { 1, 2, 3, 4 }; int n = sizeof(arr) / sizeof(int); cout << getMinDiff(arr, n); return 0;} |
Java
// Java implementation of the approachimport java.util.*; class GFG{ private static final int MAX = 100001; // Function to return the minimum// absolute difference between any// two elements of the arraystatic int getMinDiff(int arr[], int n){ // To store the frequency of each element int[] freq = new int[MAX]; for(int i = 0; i < n; i++) { freq[i] = 0; } for (int i = 0; i < n; i++) { // Update the frequency of current element freq[arr[i]]++; // If current element appears more than once // then the minimum absolute difference // will be 0 i.e. |arr[i] - arr[i]| if (freq[arr[i]] > 1) return 0; } int mn = Integer.MAX_VALUE; // Checking the distance between the nearest // two elements in the frequency array for (int i = 0; i < MAX; i++) { if (freq[i] > 0) { i++; int cnt = 1; while ((freq[i] == 0) && (i != MAX - 1)) { cnt++; i++; } mn = Math.min(cnt, mn); i--; } } // Return the minimum absolute difference return mn;} // Driver codepublic static void main(String[] args) { int arr[] = { 1, 2, 3, 4 }; int n = arr.length; System.out.println(getMinDiff(arr, n)); }} // This code is contributed by nidhi16bcs2007 |
Python3
# Python3 implementation of the approachMAX = 100001 # Function to return the minimum# absolute difference between any# two elements of the arraydef getMinDiff(arr, n): # To store the frequency of each element freq = [0 for i in range(MAX)] for i in range(n): # Update the frequency of current element freq[arr[i]] += 1 # If current element appears more than once # then the minimum absolute difference # will be 0 i.e. |arr[i] - arr[i]| if (freq[arr[i]] > 1): return 0 mn = 10**9 # Checking the distance between the nearest # two elements in the frequency array for i in range(MAX): if (freq[i] > 0): i += 1 cnt = 1 while ((freq[i] == 0) and (i != MAX - 1)): cnt += 1 i += 1 mn = min(cnt, mn) i -= 1 # Return the minimum absolute difference return mn # Driver codearr = [ 1, 2, 3, 4]n = len(arr) print(getMinDiff(arr, n)) # This code is contributed by Mohit Kumar |
C#
// C# implementation of the approach using System; class GFG { private static int MAX = 100001; // Function to return the minimum // absolute difference between any // two elements of the array static int getMinDiff(int []arr, int n) { // To store the frequency of each element int[] freq = new int[MAX]; for(int i = 0; i < n; i++) { freq[i] = 0; } for (int i = 0; i < n; i++) { // Update the frequency of current element freq[arr[i]]++; // If current element appears more than once // then the minimum absolute difference // will be 0 i.e. |arr[i] - arr[i]| if (freq[arr[i]] > 1) return 0; } int mn = int.MaxValue; // Checking the distance between the nearest // two elements in the frequency array for (int i = 0; i < MAX; i++) { if (freq[i] > 0) { i++; int cnt = 1; while ((freq[i] == 0) && (i != MAX - 1)) { cnt++; i++; } mn = Math.Min(cnt, mn); i--; } } // Return the minimum absolute difference return mn; } // Driver code public static void Main() { int []arr = { 1, 2, 3, 4 }; int n = arr.Length; Console.WriteLine(getMinDiff(arr, n)); } } // This code is contributed by AnkitRai01 |
Output:
1
Output:
1
Alternate Shorter Implementation :
Python3
# Python3 implementation of the approachimport itertools arr = [1,2,3,4]diff_list = [] # Get the combinations of numbersfor n1, n2 in list(itertools.combinations(arr, 2)): # Find the absolute difference diff_list.append(abs(n1-n2)) print(min(diff_list)) # This code is contributed by mailprakashindia |
Output:
1
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