The idea is simple everytime we see a start, we just push it to the stack. Now when we reach an end, we are guaranteed that the top of the stack is a start with the same id as the current item because all completed start/ends in between this start and end has been removed already. We just add current item timestamp – stack top timestamp + 1 to times[i].
 
So for example

[…, {0:start:3}] and item = {0:end:6} we add 6 – 3 + 1
 
However, what if there are function calls in between the start and end of the function of id 0? We can account for this by subtracting the length of the function calls in between the function id 0 whenever we complete an inner function marked by an end.
 
[…, {0:start:3}, {2:start:4}] and item = {2:end:5} so we increment times[2] by curr_length = 5 – 4 + 1 = 2 and then we subtract times[0] by curr_length as it takes up that amount of time out of the total time
 
So whenever we see an end, we have to make sure to subtract our curr_length to whatever function is enclosing it if it exists.
 
#include <iostream>
#include <vector>
#include <stack>
#include <sstream>
#include <cassert>

using namespace std;

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;
                times[item.id] += time_added;
                st.pop();

                if(!st.empty()) {
                    assert(st.top().status == "start");
                    times[st.top().id] -= time_added;
                }
            }
        }

        return times;
    }
};

Extract the log parsing logic as a inner class.
Calculate the function’s running time when encounter an “end” log entry. If current ended func has a main func still running (in the stack), substract the running time advance. So we don’t need to use a “prev” variable.
Another idea is using a field in the inner class to track the real running time for a function. I believe this way would be the most straightforward for myself.
Both methods follows the O(n) time complexiy, and O(n/2) extra space consumption.

 
Method 1
 
class Solution {
    public int[] exclusiveTime(int n, List<String> logs) {
        Deque<Log> stack = new ArrayDeque<>();
        int[] result = new int[n];
        for (String content : logs) {
            Log log = new Log(content);
            if (log.isStart) {
                stack.push(log);
            } else {
                Log top = stack.pop();
                result[top.id] += (log.time - top.time + 1);
                if (!stack.isEmpty()) {
                    result[stack.peek().id] -= (log.time - top.time + 1);
                }
            }
        }
        
        return result;
    }
    
    public static class Log {
        public int id;
        public boolean isStart;
        public int time;
        
        public Log(String content) {
            String[] strs = content.split(":");
            id = Integer.valueOf(strs[0]);
            isStart = strs[1].equals("start");
            time = Integer.valueOf(strs[2]);
        }
    }
}

 
Method 2
 
class Solution {
    public int[] exclusiveTime(int n, List<String> logs) {
        Deque<Log> stack = new ArrayDeque<>();
        int[] result = new int[n];
        int duration = 0;
        for (String content : logs) {
            Log log = new Log(content);
            if (log.isStart) {
                stack.push(log);
            } else {
                Log top = stack.pop();
                result[top.id] += (log.time - top.time + 1 - top.subDuration);
                if (!stack.isEmpty()) {
                    stack.peek().subDuration += (log.time - top.time + 1);
                }
            }
        }
        
        return result;
    }
    
    public static class Log {
        public int id;
        public boolean isStart;
        public int time;
        public int subDuration;
        
        public Log(String content) {
            String[] strs = content.split(":");
            id = Integer.valueOf(strs[0]);
            isStart = strs[1].equals("start");
            time = Integer.valueOf(strs[2]);
            subDuration = 0;
        }
    }
}
 

We examine two approaches – both will be stack based.
 
In a more conventional approach, let’s look between adjacent events, with duration time - prev_time. If we started a function, and we have a function in the background, then it was running during this time. Otherwise, we ended the function that is most recent in our stack.
 
def exclusiveTime(self, N, logs):
    ans = [0] * N
    stack = []
    prev_time = 0

    for log in logs:
        fn, typ, time = log.split(':')
        fn, time = int(fn), int(time)

        if typ == 'start':
            if stack:
                ans[stack[-1]] += time - prev_time 
            stack.append(fn)
            prev_time = time
        else:
            ans[stack.pop()] += time - prev_time + 1
            prev_time = time + 1

    return ans


 
In the second approach, we try to record the “penalty” a function takes. For example, if function 0 is running at time [1, 10], and function 1 runs at time [3, 5], then we know function 0 ran for 10 units of time, less a 3 unit penalty. The idea is this: Whenever a function completes using T time, any functions that were running in the background take a penalty of T. Here is a slow version to illustrate the idea:
 
def exclusiveTime(self, N, logs):
    ans = [0] * N
    #stack = SuperStack()
    stack = []

    for log in logs:
        fn, typ, time = log.split(':')
        fn, time = int(fn), int(time)

        if typ == 'start':
            stack.append(time)
        else:
            delta = time - stack.pop() + 1
            ans[fn] += delta
            #stack.add_across(delta)
            stack = [t+delta for t in stack] #inefficient

    return ans

 
This code already ACs, but it isn’t efficient. However, we can easily upgrade our stack to a “superstack” that supports self.add_across: addition over the whole array in constant time.
 
class SuperStack(object):
    def __init__(self):
        self.A = []
    def append(self, x):
        self.A.append([x, 0])
    def pop(self):
        x, y = self.A.pop()
        if self.A:
            self.A[-1][1] += y
        return x + y
    def add_across(self, y):
        if self.A:
            self.A[-1][1] += y