We scan in chronological order. No earlier scanned day answered it, and today is warmer, so today is its first valid answer.

The stack stores indices, because the output is a distance between days. Its temperatures are nonincreasing from bottom to top. A new temperature resolves a suffix of cooler waiting days; stop at the first equal or warmer day. Days left at the end have no future warmer day and keep zero.

What the function promises: Return a list of waiting distances, with zero where no strictly warmer future day exists.

def warmer_days(temperatures):
    answer = [0] * len(temperatures)
    waiting = []
    for today, temperature in enumerate(temperatures):
        while waiting and temperatures[waiting[-1]] < temperature:
            earlier = waiting.pop()
            answer[earlier] = today - earlier
        waiting.append(today)
    return answer

assert warmer_days([73, 71, 74]) == [2, 1, 0]
assert warmer_days([73, 73]) == [0, 0]
assert warmer_days([]) == []

Work and storage: O(n) time and O(n) extra space. Every index is pushed once and popped at most once; nested syntax does not imply quadratic work.

When to recognize it: Several waiting days can get their answer from the same new day. Keep unresolved days in temperature order.

Check before applying it: Equal temperatures are not warmer. Sorting temperatures would lose the original waiting distances.

Pop while the waiting temperature is less than or equal to today’s temperature. The equality rule follows the question, not a memorized template.