## 636. Exclusive Time of Functions

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

Each function has a unique id, start from0ton-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.

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

Thoughts:

1. Recursive call: so using stack: record current functions by id
2. For each log, check the type of fuction:
1. if it is a "start", increment the caller function execution time by querying the last element in the stack with time - prev_time
2. if it is a "end", pop the function from stack and use it to index into returned function. The total execution for popped function is time + 1 - prev_time since the questions records the end of the time for "end"

Code:

``````class Solution(object):
def exclusiveTime(self, n, logs):
"""
:type n: int
:type logs: List[str]
:rtype: List[int]
"""
funcs =  * n
stack = []
prev_time = 0

fid , typ, time = log.split(':')
fid , time = int(fid), int(time)

if typ == "start":
if stack:
funcs[stack[-1]] += time - prev_time
stack.append(fid)
prev_time = time
else:
funcs[stack.pop()] += time + 1 -prev_time # end of time t so + 1
prev_time = time + 1

return funcs
``````

Code: use stack to record time (slow)

``````class Solution(object):
def exclusiveTime(self, n, logs):
"""
:type n: int
:type logs: List[str]
:rtype: List[int]
"""
funcs , stack =  * n, []

fid , typ, time = log.split(':')
fid, time = int(fid), int(time)

if typ == 'start':
stack.append(time)
else:
d = time + 1 - stack.pop()
funcs[fid] += d
stack = [t + d for t in stack]

return funcs
``````

Code: Augmented solution for new stack class design

``````class Solution(object):
def exclusiveTime(self, n, logs):
"""
:type n: int
:type logs: List[str]
:rtype: List[int]
"""
funcs , stack =  * n, AccStack()

fid , typ, time = log.split(':')
fid, time = int(fid), int(time)

if typ == 'start':
stack.append(time)
else:
d = time + 1 - stack.pop()
funcs[fid] += d # adds up for adjustments later

return funcs

class AccStack(object):
def __init__ (self):
self.s = []
def append(self, x):
self.s.append([x, 0]) #[time, time "waiting" in calling other methods]
def pop(self):
t1, t2 = self.s.pop()
if self.s:
self.s[-1] += t2
return  t1 + t2