Daily Temperatures 25 XP Medium

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Given an array of daily temperatures, return an array where answer[i] is the number of days you have to wait after day i for a warmer temperature. If there is no future day, put 0.

temperatures = [73, 74, 75, 71, 69, 72, 76, 73]
# Output:       [1,  1,  4,  2,  1,  1,  0,  0]
# Day 0 (73°): next warmer is day 1 (74°) → wait 1 day
# Day 2 (75°): next warmer is day 6 (76°) → wait 4 days
# Day 6 (76°): no warmer day ahead → 0

This is a classic Monotonic Decreasing Stack problem:

Maintain a stack of indices (not values) where temperatures are decreasing
When you find a warmer temperature, pop all colder days from the stack
The wait time is current_index - popped_index
Time: O(n) — each index is pushed and popped at most once

Monotonic stacks are a powerful pattern — they appear in many interview problems!

📋 Instructions

Implement `daily_temperatures(temperatures)` using a monotonic stack. Test with [73, 74, 75, 71, 69, 72, 76, 73].

Use a stack of indices. For each day i, while the stack is not empty AND temps[i] > temps[stack[-1]], pop the index and set answer[popped] = i - popped. Then push i.

🧪 Test Cases

Input

result

Expected

[1, 1, 4, 2, 1, 1, 0, 0]

Test case 1

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main.py

Hi! I'm Rex 👋

Output

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