Difficulty Level: Rookie

Given a stream of numbers, print average (or mean) of the stream at every point. For example, let us consider the stream as 10, 20, 30, 40, 50, 60, …

Average of 1 numbers is 10.00 Average of 2 numbers is 15.00 Average of 3 numbers is 20.00 Average of 4 numbers is 25.00 Average of 5 numbers is 30.00 Average of 6 numbers is 35.00 ..................

To print mean of a stream, we need to find out how to find average when a new number is being added to the stream. To do this, all we need is count of numbers seen so far in the stream, previous average and new number. Let *n *be the count, *prev_avg *be the previous average and x be the new number being added. The average after including *x* number can be written as *(prev_avg*n + x)/(n+1)*.

## C++

#include <stdio.h> // Returns the new average after including x float getAvg(float prev_avg, int x, int n) { return (prev_avg * n + x) / (n + 1); } // Prints average of a stream of numbers void streamAvg(float arr[], int n) { float avg = 0; for (int i = 0; i < n; i++) { avg = getAvg(avg, arr[i], i); printf("Average of %d numbers is %f \n", i + 1, avg); } return; } // Driver program to test above functions int main() { float arr[] = { 10, 20, 30, 40, 50, 60 }; int n = sizeof(arr) / sizeof(arr[0]); streamAvg(arr, n); return 0; }

## Java

// Java program to find average // of a stream of numbers class GFG { // Returns the new average after including x static float getAvg(float prev_avg, float x, int n) { return (prev_avg * n + x) / (n + 1); } // Prints average of a stream of numbers static void streamAvg(float arr[], int n) { float avg = 0; for (int i = 0; i < n; i++) { avg = getAvg(avg, arr[i], i); System.out.printf("Average of %d numbers is %f \n", i + 1, avg); } return; } // Driver program to test above functions public static void main(String[] args) { float arr[] = { 10, 20, 30, 40, 50, 60 }; int n = arr.length; streamAvg(arr, n); } } // This code is contributed by Smitha Dinesh Semwal

Output :

Average of 1 numbers is 10.000000 Average of 2 numbers is 15.000000 Average of 3 numbers is 20.000000 Average of 4 numbers is 25.000000 Average of 5 numbers is 30.000000 Average of 6 numbers is 35.000000

The above function getAvg() can be optimized using following changes. We can avoid the use of prev_avg and number of elements by using static variables (Assuming that only this function is called for average of stream). Following is the oprimnized version.

#include <stdio.h> // Returns the new average after including x float getAvg(int x) { static int sum, n; sum += x; return (((float)sum) / ++n); } // Prints average of a stream of numbers void streamAvg(float arr[], int n) { float avg = 0; for (int i = 0; i < n; i++) { avg = getAvg(arr[i]); printf("Average of %d numbers is %f n", i + 1, avg); } return; } // Driver program to test above functions int main() { float arr[] = { 10, 20, 30, 40, 50, 60 }; int n = sizeof(arr) / sizeof(arr[0]); streamAvg(arr, n); return 0; }

Thanks to Abhijeet Deshpande for suggesting this optimized version.

**Related article:**

Program for average of an array (Iterative and Recursive)

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