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

• Difficulty Level : Medium
• Last Updated : 12 May, 2021

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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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

• Find the minimum and maximum elements 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 elements.
• 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`

## Javascript

 ``
Output:
`3`

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

My Personal Notes arrow_drop_up