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:

  1. Traverse through the array and create a hash array to store the frequency of the array elements.
  2. Now, traverse through the hash array and calculate the distance between two nearest elements.
  3. 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++

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// 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 array
int 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 code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(int);
  
    cout << getMinDiff(arr, n);
  
    return 0;
}

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Java














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// Java implementation of the approach 


import java.util.*;  


  


class GFG 


{  


  


private static final 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 = 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 code 


public 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 



















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Python3














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# Python3 implementation of the approach 


MAX = 100001


  


# Function to return the minimum 


# absolute difference between any 


# two elements of the array 


def 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 code 


arr = [ 1, 2, 3, 4] 


n = len(arr) 


  


print(getMinDiff(arr, n)) 


  


# This code is contributed by Mohit Kumar 



















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C#














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// 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 



















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Output:


1






          
          
          
          

            

    

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