Finding the average of a continuous stream of numbers

Find the average or mean of a continuously incoming stream of numbers at every point in time.

Example 1:

Input:
n = 5
arr[] = {10, 20, 30, 40, 50}
Output: 10.00 15.00 20.00 25.00 30.00 
Explanation: 
10 / 1 = 10.00
(10 + 20) / 2 = 15.00
(10 + 20 + 30) / 3 = 20.00
And so on.

Example 2:

Input:
n = 3
arr[] = {6, 7, 2}
Output: 6.00 6.50 5.00 
Explanation: 
6 / 1 = 6.00
(6 + 7) / 2 = 6.50
(6 + 7 + 2) / 3 = 5.00

method 1:

Initializes an array arr2 of size n to store the calculated averages. Initializes a variable sum to 0 to keep track of the running sum of the input numbers.

loop through the input array arr from index 0 to n-1. In each iteration, the current input number arr[i] is added to the running sum. The average of all the numbers up to the current index i is then calculated by dividing the sum by i+1 and stored in the corresponding index of the output array arr2[i].

Finally, the function returns the output array arr2 containing the calculated averages.

Java implementation:

class codemummy {
    float[] streamAvg(int[] arr, int n) {
        float[] arr2=new float[n];
        float sum=0;
        for(int i=0;i<n;i++){
            sum=sum+arr[i];
            arr2[i]=sum/(i+1);
        }
        return arr2;
    }
}

 Time Complexity: O(n)
 Auxiliary Space: O(1).

c++ implementation:

#include <vector>
using namespace std;

vector<float> streamAvg(int arr[], int n) {
    vector<float> arr2(n);
    float sum = 0.0;
    for (int i = 0; i < n; i++) {
        sum = sum + arr[i];
        arr2[i] = sum / (i + 1);
    }
    return arr2;
}

 Time Complexity: O(n)
 Auxiliary Space: O(1).

python implementation:

def streamAvg(arr, n):
    arr2 = [0] * n
    sum = 0
    for i in range(n):
        sum += arr[i]
        arr2[i] = sum / (i + 1)
    return arr2

 Time Complexity: O(n)
 Auxiliary Space: O(1).