Steady State

A medium Python interview practice problem on DataDriven. Write the Python and run it against test cases, with instant feedback.

Domain
Python
Difficulty
medium
Seniority
L4
Asked in
Gemini

Problem

Return the mean of every run of k consecutive samples in nums, each rounded to two decimals, as the smoothed series a metrics dashboard plots in sample order. When the buffer holds fewer than k samples there is nothing to smooth, so return an empty list.

Turn a jittery stream into a calm read, one step at a time.

Example 1

Input:

nums = [1,2,3,4,5], k = 3

Output:

[2,3,4]

Example 2

Input:

nums = [1,2,2], k = 3

Output:

[1.67]

Worked solution and explanation

What this really is

Strip the dashboard costume and this is a fixed-size sliding window. The windows overlap and advance one sample at a time, so [1, 2, 3, 4, 5] with k = 3 gives three means, not one. Candidates who chop the list into blocks of k return [2.0] and quietly drop the tail. Once the overlap is clear, the second tell is carrying a running sum instead of re-summing k numbers per window. Miss that and a long stream with a wide window turns O(n) into O(n * k).

The trap most people walk into

The short-buffer rule is load-bearing. Jump straight into summing the first k elements and a short input either throws or divides a partial sum by k, returning a wrong value instead of an empty list. Check the length before you touch nums[k - 1]. The rounding is part of the answer too: [1, 2, 2] must come back as 1.67, not 1.6666666666666667.

Step 1: Handle the short-input case first

If len(nums) < k, return [] immediately. This both satisfies the spec and protects the first-window sum from an index error.

Step 2: Seed the running sum

Sum the first k elements once. This is the only full summation you ever do.

Step 3: Slide one element at a time

For each later position, add the entering element and subtract the one that fell out of the window, then store the rounded mean. The sum updates in O(1).

The solution

Running sum, one pass
def moving_average(nums, k):
    if len(nums) < k:
        return []
    result = []
    window_sum = sum(nums[:k])
    result.append(round(window_sum / k, 2))
    for i in range(k, len(nums)):
        window_sum += nums[i] - nums[i - k]
        result.append(round(window_sum / k, 2))
    return result

Seed once, then adjust by the entering and leaving element per step.

Common follow-up questions