# Minimum distance between the maximum and minimum element of a given Array

Given an array A[] consisting of N elements, the task is to find the minimum distance between the minimum and the maximum element of the array.
Examples:

Input: arr[] = {3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 8, 2}
Output:
Explanation:
The minimum element(= 1) is present at indices {2, 4}
The maximum element(= 8) is present at indices {7, 10}.
The minimum distance between an occurrence of 1 and 8 is 7 – 4 = 3
Input: arr[] = {1, 3, 69}
Output:
Explanation:
The minimum element(= 1) is present at index 0.
The maximum element(= 69) is present at index 2.
Therefore, the minimum distance between them is 2.

Naive Approach:
The simplest approach to solve this problem is as follows:

• Find the minimum and maximum element of the array.
• Traverse the array and for every occurrence of the maximum element, calculate its distance from all occurrences of the minimum element in the array and update the minimum distance.
• After complete traversal of the array, print all the minimum distance obtained.

Time Complexity: O(N2
Auxiliary Space: O(1)
Efficient Approach:
Follow the steps below to optimize the above approach:

• Traverse the array to find the minimum and maximum element.
• Initialize two variables min_index and max_index to store the indices of the minimum and maximum elements of the array respectively. Initialize them with -1.
• Traverse the array. If at any instant, both min_index and max_index is not equal to -1, i.e. both of them have stored a valid index, calculate there a difference.
• Compare this difference with the minimum distance(say, min_dist) and update min_dist accordingly.
• Finally, print the final value of min_dist obtained after the complete traversal of the array.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement the` `// above approach` `#include ` `using` `namespace` `std;`   `// Function to find the minimum` `// distance between the minimum` `// and the maximum element` `int` `minDistance(``int` `a[], ``int` `n)` `{` `    ``// Stores the minimum and maximum` `    ``// array element` `    ``int` `maximum = -1, minimum = INT_MAX;`   `    ``// Stores the most recently traversed` `    ``// indices of the minimum and the` `    ``// maximum element` `    ``int` `min_index = -1, max_index = -1;`   `    ``// Stores the minimum distance` `    ``// between the minimum and the` `    ``// maximium` `    ``int` `min_dist = n + 1;`   `    ``// Find the maximum and` `    ``// the minimum element` `    ``// from the given array` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``if` `(a[i] > maximum)` `            ``maximum = a[i];`   `        ``if` `(a[i] < minimum)` `            ``minimum = a[i];` `    ``}`   `    ``// Find the minimum distance` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// Check if current element` `        ``// is equal to minimum` `        ``if` `(a[i] == minimum)` `            ``min_index = i;`   `        ``// Check if current element` `        ``// is equal to maximum` `        ``if` `(a[i] == maximum)` `            ``max_index = i;`   `        ``// If both the minimum and the` `        ``// maximum element has` `        ``// occurred at least once` `        ``if` `(min_index != -1` `            ``&& max_index != -1)`   `            ``// Update the minimum distance` `            ``min_dist` `                ``= min(min_dist,` `                      ``abs``(min_index` `                          ``- max_index));` `    ``}`   `    ``// Return the answer` `    ``return` `min_dist;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `a[] = { 3, 2, 1, 2, 1, 4,` `                ``5, 8, 6, 7, 8, 2 };` `    ``int` `n = ``sizeof` `a / ``sizeof` `a;` `    ``cout << minDistance(a, n);` `}`

## Java

 `// Java Program to implement` `// the above approach` `import` `java.util.*;` `class` `GFG {`   `    ``// Function to find the minimum` `    ``// distance between the minimum` `    ``// and the maximum element` `    ``public` `static` `int` `minDistance(``int` `a[], ``int` `n)` `    ``{`   `        ``// Stores the minimum and maximum` `        ``// array element` `        ``int` `max = -``1``, min = Integer.MAX_VALUE;`   `        ``// Stores the most recently traversed` `        ``// indices of the minimum and the` `        ``// maximum element` `        ``int` `min_index = -``1``, max_index = -``1``;`   `        ``// Stores the minimum distance` `        ``// between the minimum and the` `        ``// maximium` `        ``int` `min_dist = n + ``1``;`   `        ``// Find the maximum and` `        ``// the minimum element` `        ``// from the given array` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``if` `(a[i] > max)` `                ``max = a[i];` `            ``if` `(a[i] < min)` `                ``min = a[i];` `        ``}`   `        ``// Find the minimum distance` `        ``for` `(``int` `i = ``0``; i < n; i++) {`   `            ``// Check if current element` `            ``// is equal to minimum` `            ``if` `(a[i] == min)` `                ``min_index = i;`   `            ``// Check if current element` `            ``// is equal to maximum` `            ``if` `(a[i] == max)` `                ``max_index = i;`   `            ``// If both the minimum and the` `            ``// maximum element has` `            ``// occurred at least once` `            ``if` `(min_index != -``1` `                ``&& max_index != -``1``)` `                ``min_dist` `                    ``= Math.min(min_dist,` `                               ``Math.abs(min_index` `                                        ``- max_index));` `        ``}` `        ``return` `min_dist;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `n = ``12``;` `        ``int` `a[] = { ``3``, ``2``, ``1``, ``2``, ``1``, ``4``,` `                    ``5``, ``8``, ``6``, ``7``, ``8``, ``2` `};` `        ``System.out.println(minDistance(a, n));` `    ``}` `}`

## Python3

 `# Python3 Program to implement the` `# above approach` `import` `sys`   `# Function to find the minimum` `# distance between the minimum` `# and the maximum element` `def` `minDistance(a, n):`   `    ``# Stores the minimum and maximum` `    ``# array element` `    ``maximum ``=` `-``1` `    ``minimum ``=` `sys.maxsize` ` `  `    ``# Stores the most recently traversed` `    ``# indices of the minimum and the` `    ``# maximum element` `    ``min_index ``=` `-``1` `    ``max_index ``=` `-``1` ` `  `    ``# Stores the minimum distance` `    ``# between the minimum and the` `    ``# maximium` `    ``min_dist ``=` `n ``+` `1` ` `  `    ``# Find the maximum and` `    ``# the minimum element` `    ``# from the given array` `    ``for` `i ``in` `range` `(n):` `        ``if` `(a[i] > maximum):` `            ``maximum ``=` `a[i]`   `        ``if` `(a[i] < minimum):` `            ``minimum ``=` `a[i]` ` `  `    ``# Find the minimum distance` `    ``for` `i ``in` `range` `(n):` ` `  `        ``# Check if current element` `        ``# is equal to minimum` `        ``if` `(a[i] ``=``=` `minimum):` `            ``min_index ``=` `i` ` `  `        ``# Check if current element` `        ``# is equal to maximum` `        ``if` `(a[i] ``=``=` `maximum):` `            ``max_index ``=` `i` ` `  `        ``# If both the minimum and the` `        ``# maximum element has` `        ``# occurred at least once` `        ``if` `(min_index !``=` `-``1` `and` `            ``max_index !``=` `-``1``):` ` `  `            ``# Update the minimum distance` `            ``min_dist ``=` `(``min``(min_dist,` `                        ``abs``(min_index ``-` `                            ``max_index)))` ` `  `    ``# Return the answer` `    ``return` `min_dist` ` `  `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``a ``=` `[``3``, ``2``, ``1``, ``2``, ``1``, ``4``,` `         ``5``, ``8``, ``6``, ``7``, ``8``, ``2``]` `    ``n ``=` `len``(a)` `    ``print` `(minDistance(a, n))`   `# This code is contributed by Chitranayal`

## C#

 `// C# program to implement` `// the above approach` `using` `System;`   `class` `GFG{` `    `  `// Function to find the minimum` `// distance between the minimum` `// and the maximum element` `static` `int` `minDistance(``int` `[]a, ``int` `n)` `{` `    `  `    ``// Stores the minimum and maximum` `    ``// array element` `    ``int` `max = -1, min = Int32.MaxValue;`   `    ``// Stores the most recently traversed` `    ``// indices of the minimum and the` `    ``// maximum element` `    ``int` `min_index = -1, max_index = -1;`   `    ``// Stores the minimum distance` `    ``// between the minimum and the` `    ``// maximium` `    ``int` `min_dist = n + 1;`   `    ``// Find the maximum and` `    ``// the minimum element` `    ``// from the given array` `    ``for``(``int` `i = 0; i < n; i++)` `    ``{` `        ``if` `(a[i] > max)` `            ``max = a[i];` `        ``if` `(a[i] < min)` `            ``min = a[i];` `    ``}`   `    ``// Find the minimum distance` `    ``for``(``int` `i = 0; i < n; i++)` `    ``{` `        ``// Check if current element` `        ``// is equal to minimum` `        ``if` `(a[i] == min)` `            ``min_index = i;`   `        ``// Check if current element` `        ``// is equal to maximum` `        ``if` `(a[i] == max)` `            ``max_index = i;`   `        ``// If both the minimum and the` `        ``// maximum element has` `        ``// occurred at least once` `        ``if` `(min_index != -1 && max_index != -1)` `            ``min_dist = Math.Min(min_dist,` `                                ``Math.Abs(` `                                ``min_index - ` `                                ``max_index));` `    ``}` `    ``return` `min_dist;` `}`   `// Driver Code` `public` `static` `void` `Main()` `{` `    ``int` `n = 12;` `    ``int` `[]a = { 3, 2, 1, 2, 1, 4,` `                ``5, 8, 6, 7, 8, 2 };` `                `  `    ``Console.WriteLine(minDistance(a, n));` `}` `}`   `// This code is contributed by piyush3010`

Output:

```3

```

Time Complexity: O(N)
Auxiliary Space: O(1)

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Improved By : piyush3010, chitranayal