Sum of minimum element of all sub-sequences 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 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 subsequence int 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 code int 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 subsequence def findMinSum(arr, n): occ = n - 1 Sum = 0 for i in range (n): Sum + = arr[i] * pow ( 2 , occ) occ - = 1 return Sum # Driver code arr = [ 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 subsequence function 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 subsequence function 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 code var 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.