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: 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
"end"
log for each "start"
log.import java.util.ArrayDeque
import java.util.Deque
class Solution {
fun exclusiveTime(n: Int, logs: List<String>): IntArray {
val stack: Deque<Log> = ArrayDeque()
val result = IntArray(n)
for (content in logs) {
val log = Log(content)
if (log.isStart) {
stack.push(log)
} else {
val top = stack.pop()
val executionTime = log.time - top.time + 1
result[top.id] += executionTime - top.waitingTime
if (stack.isNotEmpty()) {
stack.peek().waitingTime += executionTime
}
}
}
return result
}
private class Log internal constructor(content: String) {
var id: Int
var isStart: Boolean
var time: Int
var waitingTime: Int
init {
val tokens = content.split(":".toRegex()).dropLastWhile { it.isEmpty() }.toTypedArray()
id = tokens[0].toInt()
isStart = tokens[1] == "start"
time = tokens[2].toInt()
waitingTime = 0
}
}
}