Knew it was a monotonic stack problem pretty fast, which felt good.
Use a monotonic decreasing stack to efficiently find the next warmer day for each temperature. Iterate through the array, and for each day, pop indices from the stack while the current temperature is warmer, recording the difference in days. This yields an O(n) time and O(n) space solution.
Pro tip: Clarify that the stack stores indices, not temperatures, to easily compute the day difference. Mention that this pattern is reusable for problems like 'next greater element'.
Restate the problem: for each day, find the number of days until a warmer temperature; if none, output 0. Confirm input/output format and edge cases.
Select a stack to keep track of indices of days with decreasing temperatures. This allows efficient retrieval of the next warmer day.
Loop through the temperature array. While the stack is not empty and the current temperature is greater than the temperature at the index on top of the stack, pop and compute the difference in indices.
After the loop, any indices left in the stack have no warmer day, so their corresponding result remains 0.
State that each index is pushed and popped at most once, giving O(n) time and O(n) space. Compare with brute-force O(n^2) to highlight efficiency.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.