Number of subarrays whose minimum and maximum are same

Given an array of n integers, find the no of subarrays whose minimal and maximum elements are the same. A subarray is defined as a non-empty sequence of consecutive elements.

Examples:

Input: 2 3 1 1 
Output: 5
Explanation: The subarrays are (2), (3), (1), (1) and (1, 1) 

Input: 2 4 5 3 3 3
Output: 9
Explanation: The subarrays are (2), (4), (5), (3), (3, 3), (3, 3, 3), (3), (3, 3) and (3) 

The first thing to observe is that only those subarrays whose all elements are same will have the same minimum and maximum. Having different elements clearly means different minimum and maximum. Hence, we just need to calculate the number of continuous same elements (say d), then by combinations’ formula we get the no of subarrays to be –

No of subarrays possible with d elements = (d * (d+1) / 2) 
where d is a number of continuous same elements. 

We traverse from 1-n and then from I+1 to n and then find the number of continuous same elements and then add to the result the no subarrays possible. 

Below is the implementation of the above approach:

C++

// CPP program to count number of subarrays
// having same minimum and maximum.
#include <bits/stdc++.h>
using namespace std;

// calculate the no of contiguous subarrays
// which has same minimum and maximum
int calculate(int a[], int n)
{
    // stores the answer
    int ans = 0;

    // loop to traverse from 0-n
    for (int i = 0; i < n; i++) {

        // start checking subarray from next element
        int r = i + 1;

        // traverse for finding subarrays
        for (int j = r; j < n; j++) {

            // if the elements are same then
            // we check further and keep a count
            // of same numbers in 'r'
            if (a[i] == a[j])
                r += 1;
            else
                break;
        }

        // the no of elements in between r and i
        // with same elements.
        int d = r - i;

        // the no of subarrays that can be formed
        // between i and r
        ans += (d * (d + 1) / 2);

        // again start checking from the next index
        i = r - 1;
    }

    // returns answer
    return ans;
}

// driver program to test the above function
int main()
{
    int a[] = { 2, 4, 5, 3, 3, 3 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << calculate(a, n);
    return 0;
}

Java

// Java program to count number of subarrays
// having same minimum and maximum.

class Subarray
{
    // calculate the no of contiguous subarrays
    // which has same minimum and maximum
    static int calculate(int a[], int n)
    {
        // stores the answer
        int ans = 0;

        // loop to traverse from 0-n
        for (int i = 0; i < n; i++) {

            // start checking subarray from
            // next element
            int r = i + 1;

            // traverse for finding subarrays
            for (int j = r; j < n; j++) {

                // if the elements are same then
                // we check further and keep a
                // count of same numbers in 'r'
                if (a[i] == a[j])
                    r += 1;
                else
                    break;
            }

            // the no of elements in between r
            // and i with same elements.
            int d = r - i;

            // the no. of subarrays that can be
            // formed between i and r
            ans += (d * (d + 1) / 2);

            // again start checking from the next
            // index
            i = r - 1;
        }

        // returns answer
        return ans;
    }

    // Driver program to test above functions
    public static void main(String[] args)
    {
    int a[] = {  2, 4, 5, 3, 3, 3 };
    System.out.println(calculate(a, a.length));
    }
}
// This code is contributed by Prerna Saini

Python3

# Python3 program to count
# number of subarrays having
# same minimum and maximum.

# calculate the no of contiguous
# subarrays which has same
# minimum and maximum
def calculate(a, n):

    # stores the answer
    ans = 0;
    i = 0;

    # loop to traverse from 0-n
    while(i < n):

        # start checking subarray
        # from next element
        r = i + 1;

        # traverse for
        # finding subarrays
        for j in range(r, n):

            # if the elements are same
            # then we check further
            # and keep a count of same
            # numbers in 'r'
            if (a[i] == a[j]):
                r = r + 1;
            else:
                break;

        # the no of elements in
        # between r and i with
        # same elements.
        d = r - i;

        # the no of subarrays that
        # can be formed between i and r
        ans = ans + (d * (d + 1) / 2);

        # again start checking
        # from the next index
        i = r - 1;
        i = i + 1;

    # returns answer
    return int(ans);

# Driver Code
a = [ 2, 4, 5, 3, 3, 3 ];
n = len(a);
print(calculate(a, n));

# This code is contributed by mits

C#

// Program to count number
// of subarrays having same
// minimum and maximum.
using System;

class Subarray {
    // calculate the no of contiguous
    // subarrays which has the same
    // minimum and maximum
    static int calculate(int[] a, int n)
    {
        // stores the answer
        int ans = 0;

        // loop to traverse from 0-n
        for (int i = 0; i < n; i++) {

            // start checking subarray
            // from next element
            int r = i + 1;

            // traverse for finding subarrays
            for (int j = r; j < n; j++) {

                // if the elements are same then
                // we check further and keep a
                // count of same numbers in 'r'
                if (a[i] == a[j])
                    r += 1;
                else
                    break;
            }

            // the no of elements in between
            // r and i with same elements.
            int d = r - i;

            // the no. of subarrays that can
            // be formed between i and r
            ans += (d * (d + 1) / 2);

            // again start checking from
            // the next index
            i = r - 1;
        }

        // returns answer
        return ans;
    }

    // Driver program
    public static void Main()
    {
        int[] a = { 2, 4, 5, 3, 3, 3 };
        Console.WriteLine(calculate(a, a.Length));
    }
}

// This code is contributed by Anant Agarwal.

PHP

<?php
// PHP program to count number
// of subarrays having same
// minimum and maximum.

// calculate the no of contiguous
// subarrays which has same minimum
// and maximum
function calculate($a, $n)
{
    // stores the answer
    $ans = 0;

    // loop to traverse from 0-n
    for ($i = 0; $i < $n; $i++)
    {

        // start checking subarray
        // from next element
        $r = $i + 1;

        // traverse for finding subarrays
        for ($j = $r; $j < $n; $j++)
        {

            // if the elements are same
            // then we check further and
            // keep a count of same numbers
            // in 'r'
            if ($a[$i] == $a[$j])
                $r += 1;
            else
                break;
        }

        // the no of elements in between
        //  r and i with same elements.
        $d = $r - $i;

        // the no of subarrays that
        // can be formed between i and r
        $ans += ($d * ($d + 1) / 2);

        // again start checking
        // from the next index
        $i = $r - 1;
    }

    // returns answer
    return $ans;
}

// Driver Code
$a = array( 2, 4, 5, 3, 3, 3 );
$n = count($a);
echo calculate($a, $n);

// This code is contributed by Sam007
?>

Javascript

<script>
// JavaScript program to count number of subarrays
// having same minimum and maximum.


    // calculate the no of contiguous subarrays
    // which has same minimum and maximum
    function calculate(a, n)
    {
        // stores the answer
        let ans = 0;

        // loop to traverse from 0-n
        for (let i = 0; i < n; i++) {

            // start checking subarray from
            // next element
            let r = i + 1;

            // traverse for finding subarrays
            for (let j = r; j < n; j++) {

                // if the elements are same then
                // we check further and keep a
                // count of same numbers in 'r'
                if (a[i] == a[j])
                    r += 1;
                else
                    break;
            }

            // the no of elements in between r
            // and i with same elements.
            let d = r - i;

            // the no. of subarrays that can be
            // formed between i and r
            ans += (d * (d + 1) / 2);

            // again start checking from the next
            // index
            i = r - 1;
        }

        // returns answer
        return ans;
    }

// Driver Code

    let a = [ 2, 4, 5, 3, 3, 3 ];
    document.write(calculate(a, a.length));

</script>

Output

9

Time Complexity: O(n), where n is the size of the given array.
Auxiliary Space: O(1)