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 2^n-1 times, the second minimum occurs 2^n-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. 2^n-2 where n = 3. 
3 occurs 1 times i.e. 2^n-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>

Output:

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.