Minimum distance between the maximum and minimum element of a given Array
Last Updated :
14 Jan, 2022
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 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
// 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 Code
int 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 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
// 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 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
# 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 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# 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
// 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()
{
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
<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>
Time Complexity: O(N)
Auxiliary Space: O(1)
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