Given a sorted array A of n integers. The task is to find the sum of the minimum of all possible subsequences of A.
Note: Considering there will be no overflow of numbers.
Examples:
Input: A = [1, 2, 4, 5]
Output: 29
Subsequences are [1], [2], [4], [5], [1, 2], [1, 4], [1, 5], [2, 4], [2, 5], [4, 5] [1, 2, 4], [1, 2, 5], [1, 4, 5], [2, 4, 5], [1, 2, 4, 5]
Minimums are 1, 2, 4, 5, 1, 1, 1, 2, 2, 4, 1, 1, 1, 2, 1.
Sum is 29
Input: A = [1, 2, 3]
Output: 11
Approach: The Naive approach is to generate all possible subsequences, 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 2n-1 times, the second minimum occurs 2n-2 times, and so on… Let’s take an example:
arr[] = {1, 2, 3}
Subsequences are {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}
Minimum of each subsequence: {1}, {2}, {3}, {1}, {1}, {2}, {1}.
where
1 occurs 4 times i.e. 2 n-1 where n = 3.
2 occurs 2 times i.e. 2n-2 where n = 3.
3 occurs 1 times i.e. 2n-3 where n = 3.
So, traverse the array and add current element i.e. arr[i]* pow(2, n-1-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 subsequenceint findMinSum(int arr[], int n){ int occ = n - 1, sum = 0; for (int i = 0; i < n; i++) { sum += arr[i] * pow(2, occ); occ--; } return sum;}// Driver codeint main(){ int arr[] = { 1, 2, 4, 5 }; 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 subsequence static int findMinSum(int arr[], int n) { int occ = n - 1, sum = 0; for (int i = 0; i < n; i++) { sum += arr[i] * (int)Math.pow(2, occ); occ--; } return sum; } // Driver code public static void main(String[] args) { int arr[] = { 1, 2, 4, 5 }; 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 subsequencedef findMinSum(arr, n): occ = n - 1 Sum = 0 for i in range(n): Sum += arr[i] * pow(2, occ) occ -= 1 return Sum# Driver codearr = [1, 2, 4, 5]n = len(arr)print(findMinSum(arr, n))# This code is contributed # by mohit kumar |
C#
// C# implementation of the above approach using System;class GFG{ // Function to find the sum // of minimum of all subsequence static int findMinSum(int []arr, int n) { int occ = n - 1, sum = 0; for (int i = 0; i < n; i++) { sum += arr[i] *(int) Math.Pow(2, occ); occ--; } return sum; } // Driver code public static void Main(String []args) { int []arr = { 1, 2, 4, 5 }; 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 subsequencefunction findMinSum($arr, $n){ $occ1 = ($n); $occ = $occ1 - 1; $Sum = 0; for ($i = 0; $i < $n; $i++) { $Sum += $arr[$i] * pow(2, $occ); $occ -= 1; } return $Sum;}// Driver code$arr = array(1, 2, 4, 5);$n = count($arr);echo findMinSum($arr, $n);// This code is contributed// by Srathore?> |
Javascript
<script>// Javascript implementation of the above approach// Function to find the sum// of minimum of all subsequencefunction findMinSum(arr, n){ var occ = n - 1, sum = 0; for (var i = 0; i < n; i++) { sum += arr[i] * Math.pow(2, occ); occ--; } return sum;}// Driver codevar arr = [ 1, 2, 4, 5 ];var n = arr.length;document.write( findMinSum(arr, n));</script> |
29
Time Complexity: O(nlogn)
Auxiliary Space: O(1)
Note: To find the Sum of maximum element of all subsequences in a sorted array, just traverse the array in reverse order and apply the same formula for Sum.
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