Exclusive Time of Functions

arraystack

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 i ^th 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 i ^th index represents the exclusive time for the function with ID i .

Examples

Example 1

Input: n = 2, logs = ["0:start:0","1:start:2","1:end:5","0:end:6"]
Output: [3,4]
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]
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]
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

2 <= logs.length <= 500

0 <= function_id < n

0 <= timestamp <= 10 ^9

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.

🤔

Medium

Difficulty

Topic Info

ModuleDSA

CategoryArray

Sub-topicStack