# Find minimum difference between any two elements | Set 2

• Difficulty Level : Easy
• Last Updated : 09 Nov, 2021

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 ` `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

 ``

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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