Problem description:

Given the running logs of n functions that are executed in a nonpreemptive single threaded CPU, find the exclusive time of these functions.

Each function has a unique id, start from 0 to n-1. A function may be called recursively or by another function.

A log is a string has this format : function_id:start_or_end:timestamp. For example, “0:start:0” means function 0 starts from the very beginning of time 0. “0:end:0” means function 0 ends to the very end of time 0.

Exclusive time of a function is defined as the time spent within this function, the time spent by calling other functions should not be considered as this function’s exclusive time. You should return the exclusive time of each function sorted by their function id.

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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 time 0, then it executes 2 units of time and reaches the end of time 1. 
Now function 0 calls function 1, function 1 starts at time 2, executes 4 units of time and end at time 5.
Function 0 is running again at time 6, and also end at the time 6, thus executes 1 unit of time. 
So function 0 totally execute 2 + 1 = 3 units of time, and function 1 totally execute 4 units of time.

Note:

  1. Input logs will be sorted by timestamp, NOT log id.
  2. Your output should be sorted by function id, which means the 0th element of your output corresponds to the exclusive time of function 0.
  3. Two functions won’t start or end at the same time.
  4. Functions could be called recursively, and will always end.
  5. 1 <= n <= 100

Solution:

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class Solution:
    def exclusiveTime(self, n: int, logs: List[str]) -> List[int]:
        res = [0 for i in range(n)]
        prev_time = 0
        stack = deque()
        for log in logs:
            id, type, time = log.split(':')
            id, time = int(id), int(time)
            if type == 'start':
                if stack:
                    res[stack[-1]] += time - prev_time
                stack.append(id)
                prev_time = time
            else:
                res[stack.pop()] += time- prev_time + 1
                prev_time = time+1
        return res

The exclusive time of function means how many times that CPU spent on this function.
example:

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["A:start:0","A:start:2","A:end:5","B:start:7","B:end:7","A:end:8"]

   time 0
        |---|---|---|---|---|---|---|---|---|---|---|---|
process A------>             -------   >A
                A---------->A           
                                    B-->B
       start   start       end   start  end

The answer would be [8,1], A has exclusive 8 slots, B has exclusive 1 slot.

With this example, we can know that if a function hasn’t end, we need to keep it until it ends. So we can use a stack to store the functions. The stack only needs to store the index of each function.

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class Solution {
public:
    vector<int> exclusiveTime(int n, vector<string>& logs) {
        vector<int> res(n, 0);
        stack<int> stk;
        int preTime= 0;
        for(auto log: logs){
            
            //extract information from log, which function, start or end, happened time
            int found1= log.find(":"); //get index of function
            int found2= log.find_last_of(":"); //time frame
            int idx= stoi(log.substr(0, found1));
            string type= log.substr(found1+1, found2-found1-1); //"start" or "end"
            int time = stoi(log.substr(found2+1));
            
            if(!stk.empty()){
                res[stk.top()]+= time-preTime;
            }
            preTime = time;
            if(type == "start") stk.push(idx);
            else{
                auto tmp= stk.top(); stk.pop();
                ++res[tmp];
                ++preTime;
            }
        }
        return res;
    }
};

time complexity: $O(n)$
space complexity: $O(n)$
reference:
https://goo.gl/63GMUn