Introduction

In this blog post, we will explore how to solve the LeetCode problem "636" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

On a single-threaded CPU, we execute a program containing n functions. Each function has a unique ID between 0 and n-1. Function calls are stored in a call stack: when a function call starts, its ID is pushed onto the stack, and when a function call ends, its ID is popped off the stack. The function whose ID is at the top of the stack is the current function being executed. Each time a function starts or ends, we write a log with the ID, whether it started or ended, and the timestamp. You are given a list logs, where logs[i] represents the ith log message formatted as a string "{function_id}:{"start" | "end"}:{timestamp}". For example, "0:start:3" means a function call with function ID 0 started at the beginning of timestamp 3, and "1:end:2" means a function call with function ID 1 ended at the end of timestamp 2. Note that a function can be called multiple times, possibly recursively. A function's exclusive time is the sum of execution times for all function calls in the program. For example, if a function is called twice, one call executing for 2 time units and another call executing for 1 time unit, the exclusive time is 2 + 1 = 3. Return the exclusive time of each function in an array, where the value at the ith index represents the exclusive time for the function with ID i. Example 1: Input: n = 2, logs = ["0:start:0","1:start:2","1:end:5","0:end:6"] Output: [3,4] Explanation: Function 0 starts at the beginning of time 0, then it executes 2 for units of time and reaches the end of time 1. Function 1 starts at the beginning of time 2, executes for 4 units of time, and ends at the end of time 5. Function 0 resumes execution at the beginning of time 6 and executes for 1 unit of time. So function 0 spends 2 + 1 = 3 units of total time executing, and function 1 spends 4 units of total time executing. Example 2: Input: n = 1, logs = ["0:start:0","0:start:2","0:end:5","0:start:6","0:end:6","0:end:7"] Output: [8] Explanation: Function 0 starts at the beginning of time 0, executes for 2 units of time, and recursively calls itself. Function 0 (recursive call) starts at the beginning of time 2 and executes for 4 units of time. Function 0 (initial call) resumes execution then immediately calls itself again. Function 0 (2nd recursive call) starts at the beginning of time 6 and executes for 1 unit of time. Function 0 (initial call) resumes execution at the beginning of time 7 and executes for 1 unit of time. So function 0 spends 2 + 4 + 1 + 1 = 8 units of total time executing. Example 3: Input: n = 2, logs = ["0:start:0","0:start:2","0:end:5","1:start:6","1:end:6","0:end:7"] Output: [7,1] Explanation: Function 0 starts at the beginning of time 0, executes for 2 units of time, and recursively calls itself. Function 0 (recursive call) starts at the beginning of time 2 and executes for 4 units of time. Function 0 (initial call) resumes execution then immediately calls function 1. Function 1 starts at the beginning of time 6, executes 1 unit of time, and ends at the end of time 6. Function 0 resumes execution at the beginning of time 6 and executes for 2 units of time. So function 0 spends 2 + 4 + 1 = 7 units of total time executing, and function 1 spends 1 unit of total time executing. Constraints: 1 <= n <= 100 1 <= logs.length <= 500 0 <= function_id < n 0 <= timestamp <= 109 No two start events will happen at the same timestamp. No two end events will happen at the same timestamp. Each function has an "end" log for each "start" log.

Explanation

Here's a breakdown of the solution:

• Stack-Based Tracking: Utilize a stack to maintain the order of function calls. When a function starts, push its ID and timestamp onto the stack. When a function ends, pop from the stack.
• Time Calculation: Upon a function's end, calculate its exclusive time by subtracting the start timestamp from the end timestamp. Account for time spent in other functions by subtracting time attributed to nested calls.

• Result Aggregation: Accumulate the exclusive time for each function in a result array.

• Runtime Complexity: O(N), where N is the number of logs. Storage Complexity: O(N)

Code

    def exclusive_time(n: int, logs: list[str]) -> list[int]:
    """
    Calculates the exclusive time of each function given a list of logs.

    Args:
        n: The number of functions.
        logs: A list of log messages.

    Returns:
        A list of exclusive times for each function.
    """

    result = [0] * n
    stack = []  # Store (function_id, start_time)
    prev_time = 0

    for log in logs:
        parts = log.split(":")
        func_id = int(parts[0])
        event_type = parts[1]
        timestamp = int(parts[2])

        if event_type == "start":
            if stack:
                result[stack[-1][0]] += timestamp - prev_time
            stack.append((func_id, timestamp))
            prev_time = timestamp
        else:  # event_type == "end"
            func_id_ended, start_time = stack.pop()
            duration = timestamp - start_time + 1
            result[func_id_ended] += duration
            prev_time = timestamp + 1

    return result