Sum of minimum element of all subarrays of a sorted array

Given a sorted array A of n integers. The task is to find the sum of the minimum of all possible subarrays of A.

Examples:  

Input: A = [ 1, 2, 4, 5] 
Output: 23 
Subsequences are [1], [2], [4], [5], [1, 2], [2, 4], [4, 5] [1, 2, 4], [2, 4, 5], [1, 2, 4, 5] 
Minimums are 1, 2, 4, 5, 1, 2, 4, 1, 2, 1. 
Sum is 23
Input: A = [1, 2, 3] 
Output: 10 

Approach: The Naive approach is to generate all possible subarrays, find their minimum and add them to the result. 
Efficient Approach: It is given that the array is sorted, so observe that the minimum element occurs N times, the second minimum occurs N-1 times, and so on... Let's take an example:
 

arr[] = {1, 2, 3} 
Subarrays are {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 2, 3} 
Minimum of each subarray: {1}, {2}, {3}, {1}, {2}, {1}. 
where 
1 occurs 3 times i.e. n times when n = 3. 
2 occurs 2 times i.e. n-1 times when n = 3. 
3 occurs 1 times i.e. n-2 times when n = 3.

So, traverse the array and add the current element i.e. (arr[i]* n-i) to the sum.
Below is the implementation of the above approach: 
 

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

// Function to find the sum
// of minimum of all subarrays
int findMinSum(int arr[], int n)
{

    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i] * (n - i);

    return sum;
}

// Driver code
int main()
{
    int arr[] = { 3, 5, 7, 8 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << findMinSum(arr, n);

    return 0;
}

// Java implementation of the above approach 
class GfG 
{

// Function to find the sum 
// of minimum of all subarrays 
static int findMinSum(int arr[], int n) 
{ 

    int sum = 0; 
    for (int i = 0; i < n; i++) 
        sum += arr[i] * (n - i); 

    return sum; 
} 

// Driver code 
public static void main(String[] args) 
{ 
    int arr[] = { 3, 5, 7, 8 }; 
    int n = arr.length; 

    System.out.println(findMinSum(arr, n)); 
}
} 

// This code is contributed by Prerna Saini

# Python3 implementation of the 
# above approach 

# Function to find the sum 
# of minimum of all subarrays 
def findMinSum(arr, n):
    sum = 0
    for i in range(0, n): 
        sum += arr[i] * (n - i) 
    return sum

# Driver code 
arr = [3, 5, 7, 8 ] 
n = len(arr)

print(findMinSum(arr, n)) 

# This code has been contributed 
# by 29AjayKumar

// C# implementation of the above approach 
using System;

class GfG 
{ 

// Function to find the sum 
// of minimum of all subarrays 
static int findMinSum(int []arr, int n) 
{ 

    int sum = 0; 
    for (int i = 0; i < n; i++) 
        sum += arr[i] * (n - i); 

    return sum; 
} 

// Driver code 
public static void Main(String []args) 
{ 
    int []arr = { 3, 5, 7, 8 }; 
    int n = arr.Length; 

    Console.WriteLine(findMinSum(arr, n)); 
} 
} 

// This code is contributed by Arnab Kundu

<?php

// PHP implementation of the above approach 
// Function to find the sum 
// of minimum of all subarrays 
function findMinSum($arr,$n) 
{ 

    $sum = 0; 
    for ($i = 0; $i < $n; $i++) 
        $sum += $arr[$i] * ($n - $i); 

    return $sum; 
} 

// Driver code 
$arr = array( 3, 5, 7, 8 ); 
$n = count($arr); 

echo findMinSum($arr, $n); 
    
// This code is contributed by Arnab Kundu
?>

<script>

// Javascript implementation of the above approach

// Function to find the sum
// of minimum of all subarrays
function findMinSum(arr, n)
{

    var sum = 0;
    for (var i = 0; i < n; i++)
        sum += arr[i] * (n - i);

    return sum;
}

// Driver code
var arr = [ 3, 5, 7, 8 ];
var n = arr.length;
document.write( findMinSum(arr, n));

</script>    

49

Time Complexity: O(n)

Auxiliary Space: O(1)

Note: To find the Sum of maximum element of all subarrays in a sorted array, just traverse the array in reverse order and apply the same formula for Sum.