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>
float getAvg(float prev_avg, int x, int n)
{
return (prev_avg * n + x) / (n + 1);
}
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
", i + 1, avg);
}
return;
}
int main()
{
float arr[] = { 10, 20, 30, 40, 50, 60 };
int n = sizeof(arr) / sizeof(arr[0]);
streamAvg(arr, n);
return 0;
}
|
Java
class GFG {
static float getAvg(float prev_avg, float x, int n)
{
return (prev_avg * n + x) / (n + 1);
}
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
",
i + 1, avg);
}
return;
}
public static void main(String[] args)
{
float arr[] = { 10, 20, 30, 40, 50, 60 };
int n = arr.length;
streamAvg(arr, n);
}
}
|
Python3
def getAvg(prev_avg, x, n):
return ((prev_avg *
n + x) /
(n + 1));
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);
arr = [10, 20, 30,
40, 50, 60];
n = len(arr);
streamAvg(arr, n);
|
C#
using System;
class GFG
{
static float getAvg(float prev_avg,
float x, int n)
{
return (prev_avg * n + x) / (n + 1);
}
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;
}
public static void Main(String[] args)
{
float[] arr = {10, 20, 30,
40, 50, 60};
int n = arr.Length;
streamAvg(arr, n);
}
}
|
PHP
<?php
function getAvg($prev_avg, $x, $n)
{
return ($prev_avg * $n + $x) /
($n + 1);
}
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,"
";
}
return;
}
$arr = array(10, 20, 30, 40, 50, 60);
$n = sizeof($arr);
streamAvg($arr, $n);
?>
|
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 <stdio.h>
float getAvg(int x)
{
static int sum, n;
sum += x;
return (((float)sum) / ++n);
}
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
", i + 1, avg);
}
return;
}
int main()
{
float arr[] = { 10, 20, 30, 40, 50, 60 };
int n = sizeof(arr) / sizeof(arr[0]);
streamAvg(arr, n);
return 0;
}
|
Java
class GFG
{
static int sum, n;
static float getAvg(int x)
{
sum += x;
return (((float)sum) / ++n);
}
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;
}
public static void main(String[] args)
{
float[] arr = new float[]{ 10, 20, 30,
40, 50, 60 };
int n = arr.length;
streamAvg(arr, n);
}
}
|
Python3
def getAvg(x, n, sum):
sum = sum + x;
return float(sum) / n;
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;
arr= [ 10, 20, 30,
40, 50, 60 ];
n = len(arr);
streamAvg(arr,n);
|
C#
using System;
class GFG
{
static int sum, n;
static float getAvg(int x)
{
sum += x;
return (((float)sum) / ++n);
}
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;
}
static int Main()
{
float[] arr = new float[]{ 10, 20, 30,
40, 50, 60 };
int n = arr.Length;
streamAvg(arr, n);
return 0;
}
}
|
PHP
<?php
function getAvg($x)
{
static $sum;
static $n;
$sum += $x;
return (((float)$sum) / ++$n);
}
function streamAvg($arr, $n)
{
for ($i = 0; $i < $n; $i++)
{
$avg = getAvg($arr[$i]);
echo "Average of " . ($i + 1) .
" numbers is ".$avg."
";
}
return;
}
$arr = array(10, 20, 30,
40, 50, 60);
$n = sizeof($arr) / sizeof($arr[0]);
streamAvg($arr, $n);
?>
|
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.
Related article:
Program for average of an array (Iterative and Recursive)
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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