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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the concentration// of the resultant mixturedouble mixtureConcentration(int n, int p[]){ double res = 0; for (int i = 0; i < n; i++) res += p[i]; res /= n; return res;}// Driver codeint main(){ int arr[] = { 0, 20, 20 }; int n = sizeof(arr) / sizeof(arr[0]); cout << mixtureConcentration(n, arr);} |
C
// C implementation of the approach#include <stdio.h>// Function to return the concentration// of the resultant mixturedouble mixtureConcentration(int n, int p[]){ double res = 0; for (int i = 0; i < n; i++) res += p[i]; res /= n; return res;}// Driver codeint main(){ int arr[] = { 0, 20, 20 }; int n = sizeof(arr) / sizeof(arr[0]); printf("%f", mixtureConcentration(n, arr));}// This code is contributed by allwink45. |
Java
// Java implementation of the approachclass GFG{ // Function to return the concentration// of the resultant mixturestatic double mixtureConcentration(int n, int []p){ double res = 0; for (int i = 0; i < n; i++) res += p[i]; res /= n; return res;}// Driver codepublic 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 |
Python3
# Python3 implementation of the approach # Function to return the concentration# of the resultant mixturedef mixtureConcentration(n, p): res = 0; for i in range(n): res += p[i]; res /= n; return res;# Driver codearr = [ 0, 20, 20 ];n = len(arr);print(round(mixtureConcentration(n, arr), 4));# This code is contributed # by chandan_jnu |
C#
// C# implementation of the approachusing System;class GFG{ // Function to return the concentration// of the resultant mixturestatic 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 codestatic void Main(){ int []arr = { 0, 20, 20 }; int n = arr.Length; Console.WriteLine(mixtureConcentration(n, arr));}}// This code is contributed by chandan_jnu |
PHP
<?php// PHP implementation of the approach // Function to return the concentration// of the resultant mixturefunction 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?> |
Javascript
<script>// Javascript implementation of the approach// Function to return the concentration// of the resultant mixturefunction mixtureConcentration(n, p){ var res = 0; for(var i = 0; i < n; i++) res += p[i]; res /= n; return res;}// Driver Codevar arr = [ 0, 20, 20 ];var n = arr.length;document.write(mixtureConcentration(n, arr).toFixed(4));// This code is contributed by Ankita saini</script> |
Output:
13.3333
Time Complexity: O(N)
Auxiliary Space: O(1)
