Concentration of juice after mixing n glasses in equal proportion

Given an array arr[] where arr[i] is the concentration of juice in ith glass. The task is to find the concentration of the resultant mixture when all the glasses are mixed in equal proportions.

Examples:

Input: arr[] = {10, 20, 30}
Output: 20

Input: arr[] = {0, 20, 20}
Output: 13.3333

Approach: Since the juices are mixed in equal proportions so the resultant concentration will be the average of all the individual concentrations. Therefore the required answer would be sum(arr) / n where n is the size of the array.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the concentration
// of the resultant mixture
double mixtureConcentration(int n, int p[])
{
    double res = 0;
    for (int i = 0; i < n; i++)
        res += p[i];
    res /= n;
    return res;
}
  
// Driver code
int main()
{
    int arr[] = { 0, 20, 20 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << mixtureConcentration(n, arr);
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
  
class GFG
{
      
// Function to return the concentration
// of the resultant mixture
static double mixtureConcentration(int n, int []p)
{
    double res = 0;
    for (int i = 0; i < n; i++)
        res += p[i];
    res /= n;
    return res;
}
  
// Driver code
public static void main (String[] args) 
{
  
    int []arr = { 0, 20, 20 };
    int n = arr.length;
    System.out.println(String.format("%.4f",
                        mixtureConcentration(n, arr)));
}
}
  
// This code is contributed by chandan_jnu

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
      
# Function to return the concentration
# of the resultant mixture
def mixtureConcentration(n, p):
  
    res = 0;
    for i in range(n):
        res += p[i];
    res /= n;
    return res;
  
# Driver code
arr = [ 0, 20, 20 ];
n = len(arr);
print(round(mixtureConcentration(n, arr), 4));
  
# This code is contributed 
# by chandan_jnu

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
  
class GFG
{
      
// Function to return the concentration
// of the resultant mixture
static double mixtureConcentration(int n, int []p)
{
    double res = 0;
    for (int i = 0; i < n; i++)
        res += p[i];
    res /= n;
    return Math.Round(res,4);
}
  
// Driver code
static void Main()
{
    int []arr = { 0, 20, 20 };
    int n = arr.Length;
    Console.WriteLine(mixtureConcentration(n, arr));
}
}
  
// This code is contributed by chandan_jnu

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of the approach
      
// Function to return the concentration
// of the resultant mixture
function mixtureConcentration($n, $p)
{
    $res = 0;
    for ($i = 0; $i < $n; $i++)
        $res += $p[$i];
    $res /= $n;
    return $res;
}
  
// Driver code
$arr = array( 0, 20, 20 );
$n = count($arr);
print(round(mixtureConcentration($n, $arr), 4));
  
// This code is contributed by chandan_jnu
?>

chevron_right


Output:

13.3333

Time Complexity: O(N)



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : Chandan_Kumar