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: 3
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: 2
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 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 <bits/stdc++.h>using namespace std;// Function to find the minimum// distance between the minimum// and the maximum elementint 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 // maximum 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 Codeint main(){ int a[] = { 3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 8, 2 }; int n = sizeof a / sizeof a[0]; cout << minDistance(a, n);} |
Java
// Java Program to implement// the above approachimport 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 // maximum 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 approachimport sys# Function to find the minimum# distance between the minimum# and the maximum elementdef 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 # maximum 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 Codeif __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 approachusing System;class GFG{ // Function to find the minimum// distance between the minimum// and the maximum elementstatic 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 // maximum 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 Codepublic 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
<script>// Javascript program to implement// the above approach // Function to find the minimum // distance between the minimum // and the maximum element function minDistance(a, n) { // Stores the minimum and maximum // array element let max = -1, min = Number.MAX_VALUE; // Stores the most recently traversed // indices of the minimum and the // maximum element let min_index = -1, max_index = -1; // Stores the minimum distance // between the minimum and the // maximum let min_dist = n + 1; // Find the maximum and // the minimum element // from the given array for (let i = 0; i < n; i++) { if (a[i] > max) max = a[i]; if (a[i] < min) min = a[i]; } // Find the minimum distance for (let 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 let n = 12; let a = [ 3, 2, 1, 2, 1, 4, 5, 8, 6, 7, 8, 2 ]; document.write(minDistance(a, n));</script> |
Output:
3
Time Complexity: O(N)
Auxiliary Space: O(1)
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