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Minimum distance between peak elements in a given array

  • Last Updated : 14 Jan, 2022

Given an array arr[], the task is to find the minimum distance between the two peak elements in the given array.

Examples:

Input: arr[] = {2, 3, 1, 2, 4, 1, 2}
Output: 2
Explanation: The peak elements in the given array are {2, 3, 1, 2, 4, 1, 2}. Hence the distance between 4 and 2 is (6 – 4) = 2 which is the minimum possible.

Input: arr[] = {1, 2}
Output: -1
Explanation: There is only one peak element in the given array.

Approach: The given problem can be solved by observing the fact that for the distance to be minimum, only the distances of the adjacent peak elements is needed to be considered. Therefore, iterate the given array and for each peak element, calculate its distance from the last found peak element whose index can be maintained in a variable idx. The minimum of all these distances is the required answer.

Below is the implementation of the above approach:

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum distance
// between two peak elements
void MinimumDistance(int arr[], int n)
{
   
    // Stores the minimum distance between
    // two peak elements
    int mn = INT_MAX;
 
    // Store the index of peak element
    int idx = -1;
 
    // Checking for the 1st elements of array
    if (arr[0] >= arr[1]) {
        idx = 0;
    }
    for (int i = 1; i < n - 1; i++) {
        if (arr[i] >= arr[i - 1]
            && arr[i] >= arr[i + 1]) {
            if (idx == -1) {
                idx = i;
            }
            else {
                mn = min(mn, i - idx);
            }
            idx = i;
        }
    }
 
    // Checking for last element of the array
    if (arr[n - 1] >= arr[n - 2] && idx != -1) {
        mn = min(mn, n - 1 - idx);
    }
 
    // If number of peak elements is less than 2
    if (mn == INT_MAX)
        cout << -1;
 
    // Print Answer
    else
        cout << mn;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 1, 2, 4, 1, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
     
    MinimumDistance(arr, n);
     
    return 0;
}
 
// This code is contributed by Samim Hossain Mondal.

Java




// Java implementation of the above approach
import java.io.*;
import java.util.*;
class GFG {
 
    // Function to find the minimum distance
    // between two peak elements
    public static void MinimumDistance(int[] arr)
    {
        // Stores the minimum distance between
        // two peak elements
        int min = Integer.MAX_VALUE;
 
        // Store the index of peak element
        int idx = -1;
        int n = arr.length;
 
        // Checking for the 1st elements of array
        if (arr[0] >= arr[1]) {
            idx = 0;
        }
        for (int i = 1; i < n - 1; i++) {
            if (arr[i] >= arr[i - 1]
                && arr[i] >= arr[i + 1]) {
                if (idx == -1) {
                    idx = i;
                }
                else {
                    min = Math.min(min, i - idx);
                }
                idx = i;
            }
        }
 
        // Checking for last element of the array
        if (arr[n - 1] >= arr[n - 2] && idx != -1) {
            min = Math.min(min, n - 1 - idx);
        }
 
        // If number of peak elements is less than 2
        if (min == Integer.MAX_VALUE)
            System.out.println(-1);
 
        // Print Answer
        else
            System.out.println(min);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 2, 3, 1, 2, 4, 1, 2 };
        MinimumDistance(arr);
    }
}

Python3




# Python implementation of the above approach
 
# Function to find the minimum distance
# between two peak elements
def MinimumDistance(arr):
   
  # Stores the minimum distance between
  # two peak elements
  less = 10**9;
 
  # Store the index of peak element
  idx = -1;
  n = len(arr);
 
  # Checking for the 1st elements of array
  if (arr[0] >= arr[1]):
    idx = 0;
 
  for i in range(1, n - 1):
    if (arr[i] >= arr[i - 1] and arr[i] >= arr[i + 1]):
      if (idx == -1):
        idx = i;
      else:
        less = min(less, i - idx);
      idx = i;
     
 
  # Checking for last element of the array
  if (arr[n - 1] >= arr[n - 2] and idx != -1):
    less = min(less, n - 1 - idx);
 
 
  # If number of peak elements is less than 2
  if (less == 10**9):
    print(-1);
 
  # Print Answer
  else:
    print(less);
 
# Driver Code
 
arr = [2, 3, 1, 2, 4, 1, 2];
MinimumDistance(arr);
 
# This code is contributed by gfgking

C#




// C# implementation of the above approach
using System;
class GFG
{
 
    // Function to find the minimum distance
    // between two peak elements
    public static void MinimumDistance(int[] arr)
    {
       
        // Stores the minimum distance between
        // two peak elements
        int min = int.MaxValue;
 
        // Store the index of peak element
        int idx = -1;
        int n = arr.Length;
 
        // Checking for the 1st elements of array
        if (arr[0] >= arr[1])
        {
            idx = 0;
        }
        for (int i = 1; i < n - 1; i++)
        {
            if (arr[i] >= arr[i - 1]
                && arr[i] >= arr[i + 1])
            {
                if (idx == -1)
                {
                    idx = i;
                }
                else
                {
                    min = Math.Min(min, i - idx);
                }
                idx = i;
            }
        }
 
        // Checking for last element of the array
        if (arr[n - 1] >= arr[n - 2] && idx != -1)
        {
            min = Math.Min(min, n - 1 - idx);
        }
 
        // If number of peak elements is less than 2
        if (min == int.MaxValue)
            Console.Write(-1);
 
        // Print Answer
        else
            Console.Write(min);
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 2, 3, 1, 2, 4, 1, 2 };
        MinimumDistance(arr);
    }
}
 
// This code is contributed by gfgking

Javascript




<script>
 
// Javascript implementation of the above approach
 
 
// Function to find the minimum distance
// between two peak elements
function MinimumDistance(arr) {
  // Stores the minimum distance between
  // two peak elements
  let min = Number.MAX_SAFE_INTEGER;
 
  // Store the index of peak element
  let idx = -1;
  let n = arr.length;
 
  // Checking for the 1st elements of array
  if (arr[0] >= arr[1]) {
    idx = 0;
  }
  for (let i = 1; i < n - 1; i++) {
    if (arr[i] >= arr[i - 1]
      && arr[i] >= arr[i + 1]) {
      if (idx == -1) {
        idx = i;
      }
      else {
        min = Math.min(min, i - idx);
      }
      idx = i;
    }
  }
 
  // Checking for last element of the array
  if (arr[n - 1] >= arr[n - 2] && idx != -1) {
    min = Math.min(min, n - 1 - idx);
  }
 
  // If number of peak elements is less than 2
  if (min == Number.MAX_SAFE_INTEGER)
    document.write(-1);
 
  // Print Answer
  else
    document.write(min);
}
 
// Driver Code
 
let arr = [2, 3, 1, 2, 4, 1, 2];
MinimumDistance(arr);
// This code is contributed by gfgking
 
</script>
Output
2

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


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