# Exclusive Time Of Functions Problem

## Description

LeetCode Problem 636.

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.

Example 1:

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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:

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Input: n = 1, logs = ["0:start:0","0:start:2","0:end:5","0:start:6","0:end:6","0:end:7"]
Output: 
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:

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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:

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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:

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

Constraints:

• 1 <= n <= 100
• 1 <= 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.

## Sample C++ Code

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struct Log {
int id;
string status;
int timestamp;
};

class Solution {
public:
vector<int> exclusiveTime(int n, vector<string>& logs) {
vector<int> times(n, 0);
stack<Log> st;

for (string log : logs) {
stringstream ss(log);
string temp, temp2, temp3;
getline(ss, temp, ':');
getline(ss, temp2, ':');
getline(ss, temp3, ':');

Log item = {stoi(temp), temp2, stoi(temp3)};
if (item.status == "start") {
st.push(item);
} else {
assert(st.top().id == item.id);

int time_added = item.timestamp - st.top().timestamp + 1;
st.pop();

if (!st.empty()) {
assert(st.top().status == "start");
}
}
}

return times;
}
};
``````