Daily Temperatures
Given an array of temperatures, return an array such that `answer[i]` is the number of days you have to wait after day $i$ to get a warmer temperature. Use a Monotonic Stack.
Why Interviewers Ask This
Airbnb asks this to evaluate your mastery of stack-based optimizations for time-sensitive problems. They specifically test if you can identify the O(n^2) brute-force trap and pivot to a Monotonic Stack solution, demonstrating your ability to optimize algorithms for real-world scalability where data volume is high.
How to Answer This Question
1. Clarify requirements: Confirm input constraints like array length and temperature ranges to determine edge cases.
2. Analyze complexity: Explicitly state that a nested loop approach yields O(n^2), which is inefficient for large datasets Airbnb handles daily.
3. Propose the pattern: Introduce the Monotonic Decreasing Stack concept as the optimal strategy to track indices of pending warmer days.
4. Walk through logic: Explain iterating backward or forward while popping elements smaller than current temperatures to find the nearest greater value.
5. Implement with comments: Write clean code focusing on index management and boundary checks, verbalizing each step to show your thought process clearly.
Key Points to Cover
- Identifying the O(n^2) brute force limitation immediately
- Correctly implementing a Monotonic Decreasing Stack
- Explaining the logic of popping indices based on temperature comparison
- Handling edge cases where no warmer day exists
- Achieving optimal O(n) time and space complexity
Sample Answer
I would start by acknowledging that a naive solution using nested loops results in O(n^2) time complexity, which isn't ideal for processing large streams of weather data efficiently. Instead, I propose using a Monotonic Stack to achieve O(n) linear time complexity. The core idea is to maintain a stack of indices where the corresponding temperatures are in decreasing order. As we iterate through the array, for each current temperature, we check if it is warmer than the temperature at the index stored at the top of the stack. If it is, we pop that index, calculate the difference between the current day and the popped index, and store that result in our answer array. We repeat this until the stack is empty or the top element represents a temperature warmer than the current one. Then, we push the current index onto the stack. For example, given [73, 74, 75], when we hit 74, we pop 73's index because 74 > 73, recording a wait of 1 day. This approach ensures we only traverse the array once, making it highly performant for Airbnb's scale. Finally, any indices remaining in the stack after the loop have no future warmer day, so we assign them zero. This demonstrates both algorithmic optimization and clear problem decomposition.
Common Mistakes to Avoid
- Using nested loops resulting in Time Limit Exceeded errors on large inputs
- Confusing the stack with storing values instead of indices, losing date context
- Forgetting to initialize the answer array with zeros for unprocessed days
- Incorrectly calculating the distance between current index and stack top
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