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Sum of minimum element of all subarrays of a sorted array

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  • Difficulty Level : Medium
  • Last Updated : 22 Jun, 2022

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++




// 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




// 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




# 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#




// 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
 
// 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
?>

Javascript




<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>

Output: 

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.
 


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