Python solution with detailed explanation

• Exclusive Time of Functions https://leetcode.com/problems/exclusive-time-of-functions/#/description

Stack Simulation

• Use a stack to store state. An item in the stack consists of item_id, event_type, and the time elapsed for that item so far.
• Maintain a clk to track time in the system.
• If the stack is empty, simply add the item to the stack. The time elapsed for a newly added entry will be zero.
• If the stack is not empty and the new item is different from the top of the stack ("0:start:0") OR the new item is same as the top of stack but with the same event type (i.e. recursion example: "0:start:0"->"0:start:2"), then increment the time elapsed for the top item in the stack and add the new item.
• If the new item is same as top of the stack but the event type is end, then pop the element and store the time elapsed in a dictionary with key as item_id.
• Example: ["0:start:0","0:start:2","0:end:5","0:start:6","0:end:6","0:end:7"]
• Time complexity is O(N) where is number of logs. Space complexity is also O(N).
``````from collections import defaultdict
class Solution:
def exclusiveTime(self, n, logs):
"""
:type n: int
:type logs: List[str]
:rtype: List[int]
"""
st, clk, result = [], 0, defaultdict(int)
ID, EVENT_TYPE, ELAPSED = 0, 1, 2
items = log.split(":")
item_id, item_type, item_ts = items[0], items[1], int(items[2])
if st and st[-1][ID] == item_id and st[-1][EVENT_TYPE] != item_type:
elapsed = item_ts+1-clk+st[-1][ELAPSED]
clk = item_ts+1
result[item_id] += elapsed
st.pop()
else:
if st:
st[-1][ELAPSED] = st[-1][ELAPSED] + item_ts-clk
st.append([item_id, item_type, 0])
clk = item_ts
return [result[str(i)] for i in range(n)]
``````

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