636. Exclusive Time of Functions
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 <= 1001 <= logs.length <= 5000 <= function_id < n0 <= 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.
Solution:
class Solution {
public int[] exclusiveTime(int n, List<String> logs) {
int[] result = new int[n];
Deque<Integer> stack = new ArrayDeque<>();
// 0, 1
int preTime = 0; // 0
// ["0:start:0","1:start:2","1:end:5","0:end:6"]
for (String l : logs){
String[] log = l.split(":");
int curTime = Integer.parseInt(log[2]);
// "0:start:0" "1:start:2"
// curtime = 2
// "1:end:5"
// curtime = 5
// "0:end:6"
// curtime 6
if (log[1].equals("start")){
if (!stack.isEmpty()){ // 0
int curPeek = stack.peek();// 0
result[curPeek] = result[curPeek] + curTime - preTime;
// result[0] = result[0] + 2 - 0 = 2
}
stack.push(Integer.parseInt(log[0])); // 1
preTime = curTime; // 2
}else{
int curpop = stack.pop(); // 1 // 0
result[curpop] = result[curpop] + curTime - preTime + 1;
// [1] = [1] + 5 - 2 + 1;
// 1 = [1] + 3 + 1 = 4
// [1] = 4
// [0] = [0] + 6 - 6 + 1;
// [0] = 2+ 1 =3
preTime = curTime + 1;
// [5] + 1 = 6
// prevLogTime means the start of next interval, so we need to +1
}
}
return result;
}
}
// TC: O(n)
// SC: O(n)