# Distance between two closest minimum

• Difficulty Level : Medium
• Last Updated : 31 May, 2021

Given an array of n integers. Find the minimum distance between any two occurrences of the minimum integer in the array.
Examples:

```Input : arr[] = {5, 1, 2, 3, 4, 1, 2, 1}
Output : 2
Explanation: The minimum element 1 occurs at
indexes: 1, 5 and 7. So the minimum
distance is 7-5 = 2.

Input : arr[] = {1, 2, 1}
Output : 2```

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Brute Force Approach: The simplest approach is to find all pair of indexes of minimum element and calculate minimum distance.
Time Complexity: O(n^2), where n is the total number of elements in the array.
Efficient Approach: An efficient approach will be to observe that distance between index j and i will always be smaller than distance between indexes k and i where, k is greater than j. That is we only have to check distance between consecutive pairs of minimum elements and not all pairs. Below is the step by step algorithm:

• Find the minimum element in the array
• Find all occurrences of minimum element in the array and insert the indexes in a new array or list or vector.
• Check if size of the list of indexes is greater than one or not, i.e. the minimum element occurs atleast twice. If not than return -1.
• Traverse the list of indexes and calculate the minimum difference between any two consecutive indexes.

Below is the implementation of above idea:

## C++

 `// CPP program to find Distance between``// two closest minimum``#include ``#include ``#include ` `using` `namespace` `std;` `// function to find Distance between``// two closest minimum``int` `findClosestMin(``int` `arr[], ``int` `n)``{``    ``int` `min = INT_MAX;` `    ``// find the min element in the array``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `(arr[i] < min)``            ``min = arr[i];` `    ``// vector to store indexes of occurrences``    ``// of minimum element in the array``    ``vector<``int``> indexes;` `    ``// store indexes of occurrences``    ``// of minimum element in the array``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `(arr[i] == min)``            ``indexes.push_back(i);` `    ``// if minimum element doesnot occurs atleast``    ``// two times, return -1.``    ``if` `(indexes.size() < 2)``        ``return` `-1;` `    ``int` `min_dist = INT_MAX;` `    ``// calculate minimum difference between``    ``// any two consecutive indexes``    ``for` `(``int` `i = 1; i < indexes.size(); i++)``        ``if` `((indexes[i] - indexes[i - 1]) < min_dist)``            ``min_dist = (indexes[i] - indexes[i - 1]);` `    ``return` `min_dist;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 5, 1, 2, 3, 4, 1, 2, 1 };``    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << findClosestMin(arr, size);``    ``return` `0;``}`

## Java

 `// Java program to find Distance between``// two closest minimum``import` `java.util.Vector;` `class` `GFG {` `// function to find Distance between``// two closest minimum``    ``static` `int` `findClosestMin(``int` `arr[], ``int` `n) {``        ``int` `min = Integer.MAX_VALUE;` `        ``// find the min element in the array``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``if` `(arr[i] < min) {``                ``min = arr[i];``            ``}``        ``}` `        ``// vector to store indexes of occurrences``        ``// of minimum element in the array``        ``Vector indexes = ``new` `Vector<>();` `        ``// store indexes of occurrences``        ``// of minimum element in the array``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``if` `(arr[i] == min) {``                ``indexes.add(i);``            ``}``        ``}` `        ``// if minimum element doesnot occurs atleast``        ``// two times, return -1.``        ``if` `(indexes.size() < ``2``) {``            ``return` `-``1``;``        ``}` `        ``int` `min_dist = Integer.MAX_VALUE;` `        ``// calculate minimum difference between``        ``// any two consecutive indexes``        ``for` `(``int` `i = ``1``; i < indexes.size(); i++) {``            ``if` `((indexes.get(i) - indexes.get(i - ``1``)) < min_dist) {``                ``min_dist = (indexes.get(i) - indexes.get(i - ``1``));``            ``}``        ``}` `        ``return` `min_dist;``    ``}` `// Driver code``    ``public` `static` `void` `main(String args[]) {``        ``int` `arr[] = {``5``, ``1``, ``2``, ``3``, ``4``, ``1``, ``2``, ``1``};``        ``int` `size = arr.length;``        ``System.out.println(findClosestMin(arr, size));``    ``}``}` `// This code is contributed by PrinciRaj19992`

## Python3

 `# Python3 program to find Distance``# between two closest minimum``import` `sys` `# function to find Distance between``# two closest minimum``def` `findClosestMin(arr, n):``    ` `    ``#assigning maximum value in python``    ``min` `=` `sys.maxsize``    ` `    ` `    ``for` `i ``in` `range``(``0``, n):``        ``if` `(arr[i] < ``min``):``            ``min` `=` `arr[i]` `    ``# list in python to store indexes``    ``# of occurrences of minimum element``    ``# in the array``    ``indexes ``=` `[]` `    ``# store indexes of occurrences``    ``# of minimum element in the array``    ``for` `i ``in` `range``(``0``, n):``        ``if` `(arr[i] ``=``=` `min``):``            ``indexes.append(i)` `    ``# if minimum element doesnot occurs``    ``#  atleast two times, return -1.``    ``if` `(``len``(indexes) < ``2``):``        ``return` `-``1` `    ``min_dist ``=` `sys.maxsize` `    ``# calculate minimum difference between``    ``# any two consecutive indexes``    ``for` `i ``in` `range``(``1``, ``len``(indexes)):``        ``if` `((indexes[i] ``-` `indexes[i ``-` `1``]) < min_dist):``            ``min_dist ``=` `(indexes[i] ``-` `indexes[i ``-` `1``]);` `    ``return` `min_dist;` `# Driver code``arr ``=` `[ ``5``, ``1``, ``2``, ``3``, ``4``, ``1``, ``2``, ``1` `]``ans ``=` `findClosestMin(arr, ``8``)``print` `(ans)` `# This code is contributed by saloni1297.`

## C#

 `    ` `// C# program to find Distance between``// two closest minimum``using` `System;``using` `System.Collections.Generic;``public` `class` `GFG {`` ` `// function to find Distance between``// two closest minimum``    ``static` `int` `findClosestMin(``int` `[]arr, ``int` `n) {``        ``int` `min = ``int``.MaxValue;`` ` `        ``// find the min element in the array``        ``for` `(``int` `i = 0; i < n; i++) {``            ``if` `(arr[i] < min) {``                ``min = arr[i];``            ``}``        ``}`` ` `        ``// vector to store indexes of occurrences``        ``// of minimum element in the array``        ``List<``int``> indexes = ``new` `List<``int``>();`` ` `        ``// store indexes of occurrences``        ``// of minimum element in the array``        ``for` `(``int` `i = 0; i < n; i++) {``            ``if` `(arr[i] == min) {``                ``indexes.Add(i);``            ``}``        ``}`` ` `        ``// if minimum element doesnot occurs atleast``        ``// two times, return -1.``        ``if` `(indexes.Count < 2) {``            ``return` `-1;``        ``}``        ``int` `min_dist = ``int``.MaxValue;`` ` `        ``// calculate minimum difference between``        ``// any two consecutive indexes``        ``for` `(``int` `i = 1; i < indexes.Count; i++) {``            ``if` `((indexes[i] - indexes[i-1]) < min_dist) {``                ``min_dist = (indexes[i] - indexes[i-1]);``            ``}``        ``}`` ` `        ``return` `min_dist;``    ``}`` ` `// Driver code``    ``public` `static` `void` `Main() {``        ``int` `[]arr = {5, 1, 2, 3, 4, 1, 2, 1};``        ``int` `size = arr.Length;``        ``Console.WriteLine(findClosestMin(arr, size));``    ``}``}`` ` `// This code is contributed by PrinciRaj19992`

## PHP

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## Javascript

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Output:

`2`

Time Complexity: O(n)
Auxiliary Space: O(n)

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