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 ithindex 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 units 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.

Example 4:

Input: n = 2, logs = ["0:start:0","0:start:2","0:end:5","1:start:7","1:end:7","0:end:8"]
Output: [8,1]

Example 5:

Input: n = 1, logs = ["0:start:0","0:end:0"]
Output: [1]

 

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.

class Solution {

public:

    vector<int> exclusiveTime(int n, vector<string>& logs) {

        // base cases handle.

        

        vector<int> ans(n, 0);

        stack<int> st;

        int pretime = 0;

        for (auto log : logs) {

            int first_idx = log.find(":");

            int idx = stoi(log.substr(0, first_idx));

            int second_idx = log.find_last_of(":");

            string op = log.substr(first_idx + 1, second_idx - first_idx - 1);

            int timestamp = stoi(log.substr(second_idx + 1));

            

            if (!st.empty()) {

                ans[st.top()] += timestamp - pretime;

            }

            pretime = timestamp;

            if (op == "start") {

                st.push(idx);

            } else {

                int top = st.top(); st.pop();

                ans[top]++;

                pretime++;

            }

        }

        return ans;

    }

};

============== 2nd time =============

    vector<int> exclusiveTime(int n, vector<string>& logs) {

        vector<int> ans(n, 0);

        stack<int> st;

        int pre = -1;

        

        for (auto log : logs) {

            int task_idx = log.find(':');

            int task = stoi(log.substr(0, task_idx));

            int second_idx = log.find_last_of(':');

            string op = log.substr(task_idx + 1, second_idx - task_idx - 1);

            int time = stoi(log.substr(second_idx + 1));

            

            if (st.empty() || pre == -1) {

                pre = time;

                st.push(task);

            } else {

                if (op == "start") {

                    ans[st.top()] += time - pre;

                    st.push(task);

                    pre = time;

                } else {

                    ans[st.top()] += time - pre + 1;

                    st.pop();

                    pre = time;

                    pre++;

                }

            }

        }

        return ans;

    }