Took me a minute to realize this is just iterating over how many 4-wheelers you can fit and checking if the remainder is divisible by 2, which it always is as long as it's non-negative.
Recognize that the number of ways to combine two-wheeled and four-wheeled vehicles to sum to a total W is the number of non-negative integer solutions to 2x + 4y = W, which simplifies to x + 2y = W/2. For each total, if W is odd, return 0; otherwise, the number of ways is floor(W/4) + 1. Precompute or compute on the fly for each query.
Pro tip: Mention that this is a classic linear Diophantine equation and that the solution count can be derived in O(1) per query, which is crucial for handling large lists efficiently. Also, clarify that vehicles are distinct types (2-wheel and 4-wheel), so order does not matter.
Restate the problem: given a total wheel count, count the number of ways to choose non-negative integers x (two-wheelers) and y (four-wheelers) such that 2x + 4y = total. Note that the order of vehicles doesn't matter, so (x, y) pairs are considered distinct only by their counts.
Write the equation 2x + 4y = W. Divide by 2 to get x + 2y = W/2. This shows that W must be even; if W is odd, there are no solutions.
For even W, let N = W/2. Then x = N - 2y. Since x >= 0, we need N - 2y >= 0 => y <= N/2. Also y >= 0. So y can be any integer from 0 to floor(N/2). Thus the number of solutions is floor(N/2) + 1 = floor(W/4) + 1.
For each total in the list, check if it's odd; if so, output 0. Otherwise, compute floor(W/4) + 1. This is O(1) per query, so the overall time is O(n) for n queries.
Discuss edge cases: W = 0 (0 ways? Actually, if W=0, then x=0,y=0 is one way, but typically total wheels > 0; clarify with interviewer). Also, if W is negative, return 0. Mention potential overflow if W is very large, but in Python it's fine.
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