Class Solution


  • public class Solution
    extends Object
    636 - Exclusive Time of Functions\. Medium 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](https://en.wikipedia.org/wiki/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:** ![](https://assets.leetcode.com/uploads/2019/04/05/diag1b.png) **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.
    • Constructor Detail

      • Solution

        public Solution()
    • Method Detail

      • exclusiveTime

        public int[] exclusiveTime​(int n,
                                   List<String> logs)