This problem is from Leetcode with a description as follows:

Given an array of integers temperatures represents the daily temperatures, return an array answer such that answer[i] is the number of days you have to wait after the ith day to get a warmer temperature. If there is no future day for which this is possible, keep answer[i] == 0 instead.

My initial solution was very hard to read and failed to use stack efficiently.

```
class Solution:
def dailyTemperatures(self, temperatures: List[int]) -> List[int]:
days = []
p = 0
size = len(temperatures)
answer = [0] * size
while p<size:
if len(days)>0:
t = temperatures[p]
day = days[len(days)-1]
if t>temperatures[day]:
answer[day]=p-day
days.pop()
else:
days.append(p)
p=p+1
else:
days.append(p)
p=p+1
return answer
```

Though it worked, its runtime only beat 23% of users with Python3. That’s ~~embracing~~ embarrassing!

Finally, I came up with this solution. (With the help of ChatGPT 🤣)

```
class Solution:
def dailyTemperatures(self, temperatures: List[int]) -> List[int]:
stack = [] # Stack to store indices of temperatures
result = [0] * len(temperatures)
for i, temp in enumerate(temperatures):
# Check if the current temperature is higher than temperatures at indices in the stack
while stack and temp > temperatures[stack[-1]]:
prev_index = stack.pop()
result[prev_index] = i - prev_index
stack.append(i) # Add the current index to the stack
return result
```

As a result, its runtime now beat 73% of users with Python3.

Why am I so bad at Leetcode? 🤔