text: """
  
  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 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:
      def exclusiveTime(self, n, logs):
  
          # Init
          fid = []
          ftime = [0] * n
          prev_ts = 0
  
          # For all logs
          for log in logs:
  
              # Split into, id, type of operation and time
              # in correct format
              id, mode, ts = log.split(":")
              id, ts = int(id), int(ts)
  
              # If a function is starting
              if mode == "start":
  
                  # If there is already a function, and the current
                  # function is starting, then calculate time of the
                  # previous function w.r.t current time
                  if fid:
                      # start_time - previous_start_time
                      ftime[fid[-1]] += ts - prev_ts
  
                  # Add the current function ID on stack,
                  # this maintains the order in which the
                  # functions are called
                  fid.append(id)
  
                  # NOTE: Here, we are storing the start time,
                  # hence no need to add 1 to it, as the
                  # current function starts at the time "ts"
                  prev_ts = ts  # start_time
  
              # If function is ending
              else:
  
                  # Pop the entry from the stack.
                  if fid:
                      fid.pop()
  
                  # Since the start and end call must be mutually
                  # inclusive, we add 1 to the difference between
                  # the end and start time. The other take on the
                  # same theory is that the current function ends
                  # at time "ts+1".
                  ftime[id] += (ts + 1) - prev_ts  # end_time - start_time
  
                  # NOTE: Here, we are storing the end time,
                  # hence add 1 to it, as the currrent function
                  # ends at time "ts+1".
                  prev_ts = ts + 1  # end_time
  
          return ftime
  
  
  if __name__ == '__main__':
      s = Solution()
      n = 2
      logs = ["0:start:0", "0:start:2", "0:end:5",
              "1:start:6", "1:end:6", "0:end:7"]
      print(s.exclusiveTime(n, logs))
  