Count subarrays which contains both the maximum and minimum array element

Given an array arr[] consisting of N distinct integers, the task is to find the number of subarrays which contains both the maximum and the minimum element from the given array.

Examples:

Input: arr[] = {1, 2, 3, 4}
Output: 1 
Explanation:
Only a single subarray {1, 2, 3, 4} consists of both the maximum (= 4) and the minimum (= 1) array elements.

Input: arr[] = {4, 1, 2, 3}
Output: 3
Explanation:
Subarrays {4, 1}  , {4, 1, 2}, {4, 1, 2, 3}  consists of both the maximum(= 4) and the minimum(= 1) array elements .

Naive Approach: The simplest approach is to first, traverse the array and find the maximum and minimum of the array and then generate all possible subarrays of the given array. For each subarray, check if it contains both the maximum and the minimum array element. For all such subarrays, increase the count by 1. Finally, print the count of such subarrays.
Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach:  Follow the steps below to optimize the above approach:

• Find the index of the maximum and minimum elements. Let i and j be the respective indices such that i < j.
• All the subarray which starts from indices up to i and ends at indices after j will contain the maximum as well as the minimum array element.
• Therefore, the possible indices for the starting index of the subarray are [0, i] (total = i + 1 ).
• Therefore, the possible indices for the ending index of the subarray are [j, N – 1]  (total = N – j).
• Therefore, the count of subarrays is given by (i + 1) * ( N – j).

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count subarray
// containing  both maximum and
// minimum array elements
int countSubArray(int arr[], int n)
{
    // If the length of the
    // array is less than 2
    if (n < 2)
        return n;

    // Find the index of maximum element
    int i
        = max_element(arr, arr + n) - arr;

    // Find the index of minimum element
    int j
        = min_element(arr, arr + n) - arr;

    // If i > j, then swap
    // the value of i and j
    if (i > j)
        swap(i, j);

    // Return the answer
    return (i + 1) * (n - j);
}

// Driver Code
int main()
{
    int arr[] = { 4, 1, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    // Function call
    cout << countSubArray(arr, n);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
import java.lang.*;

class GFG{

// Function to count subarray
// containing both maximum and
// minimum array elements
static int countSubArray(int arr[], int n)
{

    // If the length of the
    // array is less than 2
    if (n < 2)
        return n;

    // Find the index of maximum element
    int i = max_element(arr);

    // Find the index of minimum element
    int j = min_element(arr);

    // If i > j, then swap
    // the value of i and j
    if (i > j)
    {
        int tmp = arr[i];
        arr[i] = arr[j];
        arr[j] = tmp;
    }

    // Return the answer
    return (i + 1) * (n - j);
}

// Function to return max_element index
static int max_element(int[] arr)
{
    int idx = 0;
    int max = arr[0];
    for(int i = 1; i < arr.length; i++)
    {
        if(max < arr[i])
        {
            max = arr[i];
            idx = i;
        }
    }
    return idx;
}

// Function to return min_element index
static int min_element(int[] arr)
{
    int idx = 0;
    int min = arr[0];
    for(int i = 1; i < arr.length; i++)
    {
        if (arr[i] < min)
        {
            min = arr[i];
            idx = i;
        }
    }
    return idx;
}

// Driver Code
public static void main (String[] args)
{
    int arr[] = { 4, 1, 2, 3 };
    int n = arr.length;

    // Function call
    System.out.println(countSubArray(arr, n));
}
}

// This code is contributed by offbeat

Python3

# Python3 program for
# the above approach

# Function to count subarray
# containing both maximum and
# minimum array elements
def countSubArray(arr, n):

    # If the length of the
    # array is less than 2
    if (n < 2):
        return n;

    # Find the index of
    # maximum element
    i = max_element(arr);

    # Find the index of
    # minimum element
    j = min_element(arr);

    # If i > j, then swap
    # the value of i and j
    if (i > j):
        tmp = arr[i];
        arr[i] = arr[j];
        arr[j] = tmp;

    # Return the answer
    return (i + 1) * (n - j);

# Function to return
# max_element index
def max_element(arr):
    idx = 0;
    max = arr[0];

    for i in range(1, len(arr)):
        if (max < arr[i]):
            max = arr[i];
            idx = i;
    return idx;

# Function to return
# min_element index
def min_element(arr):
    idx = 0;
    min = arr[0];

    for i in range(1, len(arr)):
        if (arr[i] < min):
            min = arr[i];
            idx = i;

    return idx;

# Driver Code
if __name__ == '__main__':
    arr = [4, 1, 2, 3];
    n = len(arr);

    # Function call
    print(countSubArray(arr, n));

# This code is contributed by Rajput-Ji

C#

// C# program for
// the above approach
using System;
class GFG{

// Function to count subarray
// containing both maximum and
// minimum array elements
static int countSubArray(int []arr,
                         int n)
{
  // If the length of the
  // array is less than 2
  if (n < 2)
    return n;

  // Find the index of maximum element
  int i = max_element(arr);

  // Find the index of minimum element
  int j = min_element(arr);

  // If i > j, then swap
  // the value of i and j
  if (i > j)
  {
    int tmp = arr[i];
    arr[i] = arr[j];
    arr[j] = tmp;
  }

  // Return the answer
  return (i + 1) * (n - j);
}

// Function to return max_element index
static int max_element(int[] arr)
{
  int idx = 0;
  int max = arr[0];
  for(int i = 1; i < arr.Length; i++)
  {
    if(max < arr[i])
    {
      max = arr[i];
      idx = i;
    }
  }
  return idx;
}

// Function to return min_element index
static int min_element(int[] arr)
{
  int idx = 0;
  int min = arr[0];
  for(int i = 1; i < arr.Length; i++)
  {
    if (arr[i] < min)
    {
      min = arr[i];
      idx = i;
    }
  }
  return idx;
}

// Driver Code
public static void Main(String[] args)
{
  int []arr = {4, 1, 2, 3};
  int n = arr.Length;

  // Function call
  Console.WriteLine(countSubArray(arr, n));
}
}

// This code is contributed by shikhasingrajput

Javascript

<script>
// javascript program for the
// above approach

// Function to count subarray
// containing both maximum and
// minimum array elements
function countSubArray(arr, n)
{

    // If the length of the
    // array is less than 2
    if (n < 2)
        return n;

    // Find the index of maximum element
    let i = max_element(arr);

    // Find the index of minimum element
    let j = min_element(arr);

    // If i > j, then swap
    // the value of i and j
    if (i > j)
    {
        let tmp = arr[i];
        arr[i] = arr[j];
        arr[j] = tmp;
    }

    // Return the answer
    return (i + 1) * (n - j);
}

// Function to return max_element index
function max_element(arr)
{
    let idx = 0;
    let max = arr[0];
    for(let i = 1; i < arr.length; i++)
    {
        if(max < arr[i])
        {
            max = arr[i];
            idx = i;
        }
    }
    return idx;
}

// Function to return min_element index
function min_element(arr)
{
    let idx = 0;
    let min = arr[0];
    for(let i = 1; i < arr.length; i++)
    {
        if (arr[i] < min)
        {
            min = arr[i];
            idx = i;
        }
    }
    return idx;
}

// Driver Code

     let arr = [ 4, 1, 2, 3 ];
    let n = arr.length;

    // Function call
    document.write(countSubArray(arr, n));

 // This code is contributed by avijitmondal1998
</script>

Output:

3

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