Average of a stream of numbers

Difficulty Level : Easy
Last Updated : 13 Jun, 2019

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
  ..................

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 

Python3

# Returns the new average 
# after including x 
def getAvg(prev_avg, x, n): 
    return ((prev_avg * 
             n + x) / 
            (n + 1)); 
  
# Prints average of  
# a stream of numbers 
def streamAvg(arr, n): 
    avg = 0; 
    for i in range(n): 
        avg = getAvg(avg, arr[i], i); 
        print("Average of ", i + 1, 
              " numbers is ", avg); 
  
# Driver Code 
arr = [10, 20, 30, 
       40, 50, 60]; 
n = len(arr); 
streamAvg(arr, n); 
  
# This code is contributed 
# by mits 

C#

// C# program to find average  
// of a stream of numbers 
using System; 
  
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); 
            Console.WriteLine("Average of {0} " +  
                                "numbers is {1}",  
                                     i + 1, avg); 
        } 
        return; 
    } 
  
    // Driver Code 
    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 mits 

PHP

<?php 
// PHP program for Average of 
// a stream of numbers 
  
// Returns the new average 
// after including x 
function getAvg($prev_avg, $x, $n) 
{ 
    return ($prev_avg * $n + $x) /  
                        ($n + 1); 
} 
  
// Prints average of a  
// stream of numbers 
function streamAvg($arr, $n) 
{ 
    $avg = 0; 
    for ($i = 0; $i < $n; $i++) 
    { 
        $avg = getAvg($avg, $arr[$i], $i); 
        echo "Average of ",$i + 1, "numbers is "
                                    ,$avg,"\n"; 
    } 
    return; 
} 
  
    // Driver Code 
    $arr = array(10, 20, 30, 40, 50, 60); 
    $n = sizeof($arr); 
    streamAvg($arr, $n); 
  
// This code is contributed by aj_36 
?> 

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.

C++

#include <bits/stdc++.h> 
using namespace std; 
  
// 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]);  
        cout<<"Average of "<<i+1<<" numbers is "<<fixed<<setprecision(1)<<avg<<endl;  
    }  
    return;  
}  
  
// Driver code  
int main()  
{  
    float arr[] = { 10, 20, 30, 40, 50, 60 };  
    int n = sizeof(arr) / sizeof(arr[0]);  
    streamAvg(arr, n);  
  
    return 0;  
}  
  
// This code is contributed by rathbhupendra 

C

#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; 
} 

Java

// Java program to return  
// Average of a stream of numbers 
class GFG 
{ 
static int sum, n; 
      
// Returns the new average 
// after including x 
static float getAvg(int x) 
{ 
    sum += x; 
    return (((float)sum) / ++n); 
} 
  
// 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((int)arr[i]); 
        System.out.println("Average of "+ (i + 1) +  
                           " numbers is " + avg); 
    } 
    return; 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    float[] arr = new float[]{ 10, 20, 30, 
                               40, 50, 60 }; 
    int n = arr.length; 
    streamAvg(arr, n); 
} 
} 
  
// This code is contributed by mits 

Python3

# Returns the new average 
# after including x 
def getAvg(x, n, sum): 
    sum = sum + x; 
    return float(sum) / n; 
  
# Prints average of a  
# stream of numbers 
def streamAvg(arr, n): 
    avg = 0; 
    sum = 0; 
    for i in range(n): 
        avg = getAvg(arr[i], i + 1, sum); 
        sum = avg * (i + 1); 
        print("Average of ", end = ""); 
        print(i + 1, end = ""); 
        print(" numbers is ", end = ""); 
        print(avg); 
    return; 
  
# Driver Code 
arr= [ 10, 20, 30,  
       40, 50, 60 ]; 
n = len(arr); 
streamAvg(arr,n); 
  
# This code is contributed by mits 

C#

using System; 
  
class GFG 
{ 
static int sum, n; 
      
// Returns the new average 
// after including x 
static float getAvg(int x) 
{ 
  
    sum += x; 
    return (((float)sum) / ++n); 
} 
  
// 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((int)arr[i]); 
        Console.WriteLine("Average of {0} numbers " +  
                             "is {1}", (i + 1), avg); 
    } 
    return; 
} 
  
// Driver Code 
static int Main() 
{ 
    float[] arr = new float[]{ 10, 20, 30, 
                               40, 50, 60 }; 
    int n = arr.Length; 
    streamAvg(arr, n); 
  
    return 0; 
} 
} 
  
// This code is contributed by mits 

PHP

<?php 
// Returns the new average 
// after including x 
function getAvg($x) 
{ 
    static $sum; 
    static $n; 
    $sum += $x; 
    return (((float)$sum) / ++$n); 
} 
  
// Prints average of  
// a stream of numbers 
function streamAvg($arr, $n) 
{ 
    for ($i = 0; $i < $n; $i++)  
    { 
        $avg = getAvg($arr[$i]); 
        echo "Average of " . ($i + 1) . 
             " numbers is ".$avg." \n"; 
    } 
    return; 
} 
  
// Driver Code 
$arr = array(10, 20, 30,  
             40, 50, 60); 
$n = sizeof($arr) / sizeof($arr[0]); 
streamAvg($arr, $n); 
  
// This code is contributed by mits 
?> 

Output:

Average of 1 numbers is 10.0
Average of 2 numbers is 15.0
Average of 3 numbers is 20.0
Average of 4 numbers is 25.0
Average of 5 numbers is 30.0
Average of 6 numbers is 35.0

Thanks to Abhijeet Deshpande for suggesting this optimized version.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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