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Java Solution for LeetCode algorithm problems, continually updating
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:**
**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.