Exclusive Time of Functions LeetCode solution in Python

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# Runtime: 78 ms (Top 92.98%) | Memory: 14.2 MB (Top 33.69%)
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class Solution:
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    # T=O(n), S=O(d)
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    # n=len of logs, d=depth of stack
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    def exclusiveTime(self, n: int, logs: List[str]) -> List[int]:
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        # init result array to zeroes of length n (function ids)
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        res = [0] * n
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        stack = []
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        # iterate through logs
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        for log in logs:
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            # split the log
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            # function_id: start|end: timestamp
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            log = log.split(":")
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            # type cast function id and timestamp to int type
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            id = int(log[0])
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            timestamp = int(log[2])
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            state = log[1]
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            # detect start of a function call
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            # stack[function_id, start_timestamp]
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            if state == "start":
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                # stack is non empty
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                if stack:
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                    # get the time taken by last function so far
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                    res[stack[-1][0]] += timestamp - stack[-1][1]
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                # append the current function_id and start timestamp to the stack
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                stack.append([id, timestamp])
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            else:
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                # get the time consumed by current function
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                # dont forget to add 1 as the last unit of time should be included
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                res[id] += timestamp - stack.pop()[1] + 1
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                if stack:
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                    # update the start time of last function in stack to get the cumulative result
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                    stack[-1][1] = timestamp + 1
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        return res