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++ 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);
} |
// C implementation of the approach #include <stdio.h> // 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]);
printf ( "%f" , mixtureConcentration(n, arr));
} // This code is contributed by allwink45. |
// 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 |
# 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 |
// 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 |
<?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 ?> |
<script> // Javascript implementation of the approach // Function to return the concentration // of the resultant mixture function mixtureConcentration(n, p)
{ var res = 0;
for ( var i = 0; i < n; i++)
res += p[i];
res /= n;
return res;
} // Driver Code var arr = [ 0, 20, 20 ];
var n = arr.length;
document.write(mixtureConcentration(n, arr).toFixed(4)); // This code is contributed by Ankita saini </script> |
13.3333
Time Complexity: O(N)
Auxiliary Space: O(1)