# 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++

 `// 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 `

Output:

```1
```
Output:

```1
```

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
``` My Personal Notes arrow_drop_up Front-end Developer by heart and Competitive Coder by mind

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