# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ implementation of the approach #include 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 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 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# 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 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

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :