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 subarraysint findMinSum(int arr[], int n){ int sum = 0; for (int i = 0; i < n; i++) sum += arr[i] * (n - i); return sum;}// Driver codeint 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 subarraysfunction findMinSum(arr, n){ var sum = 0; for (var i = 0; i < n; i++) sum += arr[i] * (n - i); return sum;}// Driver codevar 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.
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