To solve this problem, we can use a stack to keep track of the indices of the "unresolved" temperatures, i.e., temperatures that are waiting for a warmer day.

Here are the steps:

1. Initialize an empty stack and an array `answer` with the same length as `temperatures`, filled with 0s.

2. Iterate over the `temperatures` array from left to right. For each temperature, while the stack is not empty and the current temperature is greater than the temperature at the index at the top of the stack, pop the index from the stack and update the corresponding element in the `answer` array with the difference between the current index and the popped index.

3. After checking for and resolving any "unresolved" temperatures, push the current index onto the stack.

4. Repeat steps 2-3 for each temperature in the `temperatures` array.

5. At the end of the iteration, any indices remaining in the stack correspond to temperatures for which there is no warmer day. The corresponding elements in the `answer` array should remain 0.

Here is a Python solution following these steps:

```python
def dailyTemperatures(temperatures):
stack = []
answer = [0] * len(temperatures)
for i, temp in enumerate(temperatures):
while stack and temp temperatures[stack[-1]]:
prev_index = stack.pop()
answer[prev_index] = i - prev_index
stack.append(i)
return answer
```

Question

To solve this problem, we can use a stack to keep track of the indices of the "unresolved" temperatures, i.e., temperatures that are waiting for a warmer day.

Here are the steps:

1. Initialize an empty stack and an array `answer` with the same length as `temperatures`, filled with 0s.

2. Iterate over the `temperatures` array from left to right. For each temperature, while the stack is not empty and the current temperature is greater than the temperature at the index at the top of the stack, pop the index from the stack and update the corresponding element in the `answer` array with the difference between the current index and the popped index.

3. After checking for and resolving any "unresolved" temperatures, push the current index onto the stack.

4. Repeat steps 2-3 for each temperature in the `temperatures` array.

5. At the end of the iteration, any indices remaining in the stack correspond to temperatures for which there is no warmer day. The corresponding elements in the `answer` array should remain 0.

Here is a Python solution following these steps:

```python
def dailyTemperatures(temperatures):
    stack = []
    answer = [0] * len(temperatures)
    for i, temp in enumerate(temperatures):
        while stack and temp > temperatures[stack[-1]]:
            prev_index = stack.pop()
            answer[prev_index] = i - prev_index
        stack.append(i)
    return answer
```

Knowee AI · Accepted Answer

To solve this problem, we can use a stack to keep track of the indices of the "unresolved" temperatures, i.e., temperatures that are waiting for a warmer day.

Here are the steps:

1. Initialize an empty stack and an array `answer` with the same length as `temperatures`, filled with 0s.

2. Iterate over the `temperatures` array from left to right. For each temperature, while the stack is not empty and the current temperature is greater than the temperature at the index at the top of the stack, pop the index from the stack and update the corresponding element in the `answer` array with the difference between the current index and the popped index.

3. After checking for and resolving any "unresolved" temperatures, push the current index onto the stack.

4. Repeat steps 2-3 for each temperature in the `temperatures` array.

5. At the end of the iteration, any indices remaining in the stack correspond to temperatures for which there is no warmer day. The corresponding elements in the `answer` array should remain 0.

Here is a Python solution following these steps:

```python
def dailyTemperatures(temperatures):
    stack = []
    answer = [0] * len(temperatures)
    for i, temp in enumerate(temperatures):
        while stack and temp > temperatures[stack[-1]]:
            prev_index = stack.pop()
            answer[prev_index] = i - prev_index
        stack.append(i)
    return answer
```

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