Compute variance of a list in Python

Task

Given a Python list of numbers (ints/floats), write code to compute its variance.

Requirements

• Input: nums: list[float] (length $n\ge 1$)
• Clarify whether you are computing:
  • Population variance: $\sigma^2 = \frac{1}{n}\sum_{i=1}^n (x_i-\bar{x})^2$, or
  • Sample variance: $s^2 = \frac{1}{n-1}\sum_{i=1}^n (x_i-\bar{x})^2$ (requires $n\ge 2$)
• Avoid using numpy/pandas unless explicitly allowed.
• State time and space complexity.

Follow-ups (optional)

• Implement a numerically stable one-pass version.
• Handle edge cases (empty list, single element, very large numbers).

Compute Variance of a List

Given a list of numbers nums (length n >= 1), compute and return its population variance as a float:

variance = (1/n) * sum((x_i - mean)^2) for i = 1..n, where mean = (1/n) * sum(x_i)

Do not use numpy or pandas. For a single-element list the variance is 0. Return the result as a floating-point number.

Examples:

• nums = [2, 4, 6, 8] -> mean = 5, variance = (9+1+1+9)/4 = 5.0
• nums = [1, 1, 1, 1] -> 0.0
• nums = [10, 20] -> mean = 15, variance = (25+25)/2 = 25.0

Constraints

• 1 <= n <= 10^6
• Each element is an int or float.
• Use population variance (divide by n), not sample variance.
• Do not use numpy or pandas.

Hints

1. First compute the mean = sum(nums) / n.
2. Then sum the squared differences (x - mean)^2 over all elements.
3. Divide that sum by n (population variance). For n == 1 the result is 0.