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.