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# 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:
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(n)
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
Auxiliary Space: O(MAX)
Explanation: We have taken a frequency array of size MAX.

Alternate Shorter Implementation :

## C++

 `#include` `#include` `#include` `#include`   `using` `namespace` `std;`   `int` `main() {`   `    ``vector<``int``> arr = {1, 2, 3, 4};` `    ``vector<``int``> diff_list;`   `    ``// Get the combinations of numbers` `    ``for``(``int` `i = 0; i < arr.size(); i++) {` `        ``for``(``int` `j = i+1; j < arr.size(); j++) {`   `            ``// Find the absolute difference` `            ``diff_list.push_back(``abs``(arr[i] - arr[j]));` `        ``}` `    ``}`   `    ``// Find the minimum difference` `    ``int` `min_diff = *min_element(diff_list.begin(), diff_list.end());` `    ``cout << min_diff << endl;`   `    ``return` `0;` `}`

## 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`

## Javascript

 `let arr = [1, 2, 3, 4];` `let diffList = [];`   `// Get the combinations of numbers` `for` `(let i = 0; i < arr.length; i++) {` `    ``for` `(let j = i + 1; j < arr.length; j++) {`   `        ``// Find the absolute difference` `        ``diffList.push(Math.abs(arr[i] - arr[j]));` `    ``}` `}`   `// Find the minimum difference` `let minDiff = Math.min(...diffList);` `console.log(minDiff);`

## Java

 `import` `java.util.ArrayList;` `import` `java.util.Collections;`   `public` `class` `Main {` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``ArrayList arr = ``new` `ArrayList();` `        ``arr.add(``1``);` `        ``arr.add(``2``);` `        ``arr.add(``3``);` `        ``arr.add(``4``);` `        ``ArrayList diff_list = ``new` `ArrayList<` `            ``Integer>(); ``// Get the combinations of numbers` `        ``for` `(``int` `i = ``0``; i < arr.size(); i++) {` `            ``for` `(``int` `j = i + ``1``; j < arr.size(); j++) {`   `                ``// Find the absolute difference` `                ``diff_list.add(` `                    ``Math.abs(arr.get(i) - arr.get(j)));` `            ``}` `        ``}`   `        ``// Find the minimum difference` `        ``int` `min_diff = Collections.min(diff_list);` `        ``System.out.println(min_diff);` `    ``}` `}`

## C#

 `using` `System;` `using` `System.Collections.Generic;` `using` `System.Linq;`   `class` `Program {` `    ``static` `void` `Main(``string``[] args)` `    ``{` `        ``List<``int``> arr = ``new` `List<``int``>{ 1, 2, 3, 4 };` `        ``List<``int``> diff_list = ``new` `List<``int``>();`   `        ``// Get the combinations of numbers` `        ``for` `(``int` `i = 0; i < arr.Count; i++) {` `            ``for` `(``int` `j = i + 1; j < arr.Count; j++) {` `                ``// Find the absolute difference` `                ``diff_list.Add(Math.Abs(arr[i] - arr[j]));` `            ``}` `        ``}`   `        ``// Find the minimum difference` `        ``int` `min_diff = diff_list.Min();` `        ``Console.WriteLine(min_diff);` `    ``}` `}`

Output:

`1`

Time  Complexity: O (r* ( n C ) )
Explanation: Here r is 2 because we are making combinations of two elements using the iterator tool and n is the length of the given array, so if we calculate it, it will be O (n*(n-1) which will be O (n*n)
Auxiliary Space: O ( ( n C r  ) )  ( Here, r=2)
Explanation: We have used the list  to store all the combinations of arrays and we have made diff_list array to store absolute difference

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