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Find minimum difference between any two elements | Set 2

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  • Difficulty Level : Easy
  • Last Updated : 09 Nov, 2021
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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:
The possible absolute differences are: 
{1, 2, 3, 1, 2, 1}
Input: arr[] = {10, 2, 5, 4} 
Output:
 

 

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 the 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++




// 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;
}

Java




// 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

Python3




# 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

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

Javascript




<script>
 
// JavaScript implementation of the approach
 
 
let MAX = 100001;
 
// Function to return the minimum
// absolute difference between any
// two elements of the array
function getMinDiff(arr,n)
{
    // To store the frequency of each element
    let freq=[];
    for(let i = 0; i < n; i++)
    {
        freq[i] = 0;
    }
    for (let 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;
    }
 
    let mn = Number.MAX_VALUE;
 
    // Checking the distance between the nearest
    // two elements in the frequency array
    for (let i = 0; i < MAX; i++)
    {
        if (freq[i] > 0)
        {
            i++;
            let 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
 
    let arr = [ 1, 2, 3, 4 ];
    let n = arr.length;
     
    document.write(getMinDiff(arr, n));
 
//contributed by 171fa07058
     
</script>

Output: 

1

 

Time Complexity : O(MAX) 

Explanation : In  the above case we have taken MAX is equal to 100001, but we can take max as a maximum element in the array 

Space Complexity : O(MAX)

Explanation : We have taken frequency array of size MAX.

Alternate Shorter Implementation : 
 

Python3




# Python3 implementation of the approach
import itertools
  
arr = [1,2,3,4]
diff_list = []
  
# Get the combinations of numbers
for 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

 

Time  Complexity : O (r* ( n C ) ) 

Explanation : Here r is 2 because we are making combinations of two elements using iterator tool and n is length of given array, so if we calculate it , it will be

 O (n*(n-1) which will be O (n*n)

Space Complexity  : O ( ( n C r  ) )  ( Here, r=2)

Explanation : We have used list  to store all the combination of array and we have made diff_list array to store absolute difference


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