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++
// 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> |
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.