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 minimum of all possible subsequences of A.

Note: Considering there will be no overflow of numbers.

Examples:



Input: A = [ 1, 2, 4, 5]
Output: 17
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 result.

Efficient Approach: It is given that the array is sorted, so observe that 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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

29

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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