Efficient Polynomial & Moving Avg

This question introduces a classic algorithmic task: evaluating a polynomial with many terms at a given input value, and then relating that perspective to computing a moving average over a sliding window of data. Instead of focusing on any particular application domain, it asks you to think about how to turn a seemingly brute-force numerical procedure into something that runs in a time proportional to the number of terms or data points, even when the expressions or windows look large.

To answer it well, you need comfort with recognizing patterns in algebraic expressions and translating them into iterative algorithms. It leans heavily on ideas like rewriting expressions to expose recurrence relations, using running totals, and avoiding recomputation of overlapping work. Strong candidates typically express the algorithm clearly in words and pseudocode, reason correctly about its time complexity, and discuss edge cases. Interviewers also look for an understanding of numerical stability and implementation details, such as how to structure loops and updates so that the method is both efficient and easy to maintain.

What it tests

Many computational problems that appear to require repeated, expensive calculations can be dramatically simplified by exploiting the structure of their operations—specifically, by recognizing opportunities for incremental computation. In polynomial evaluation, the recursive nesting of terms allows us to factor out repeated multiplications, turning a seemingly quadratic process into a linear one. This is possible because each higher-degree term can be built directly from the result of the previous, lower-degree computation, leveraging the distributive property of multiplication over addition. The same incremental logic applies to moving averages: by noticing that consecutive averages share most of their terms, we can update the sum efficiently by adding the new element and subtracting the element that just left the window. The key is to identify overlapping subcomputations and find a recurrence relation that lets each step build on the last, rather than starting from scratch.