How do you solve Exclusive Time of Functions on LeetCode?

Exclusive Time of Functions (LeetCode #636) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the stack structure is crucial for managing nested function calls.

What is the brute force solution for Exclusive Time of Functions?

The brute force approach for Exclusive Time of Functions is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves iterating through each log entry and calculating the time spent on each function by checking the start and end times. This is straightforward but inefficient as it requires nested iterations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Exclusive Time of Functions?

Exclusive Time of Functions uses the following concepts: Array, Stack. The recommended patterns to study are: Stack, Array.

What are the interview tips for Exclusive Time of Functions?

Clarify the problem requirements and constraints before jumping into coding. Think aloud while coding to demonstrate your thought process. Test your code with edge cases, such as no logs or all functions starting and ending without overlap.

What are common mistakes when solving Exclusive Time of Functions?

Failing to account for the time spent in nested function calls. Incorrectly updating the previous time variable, leading to wrong calculations.

#636

Exclusive Time of Functions

Medium

ArrayStackStackArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution utilizes a stack to keep track of active function calls and calculates the exclusive time efficiently by maintaining a current time variable. This approach allows us to handle nested function calls seamlessly.

⚙️

Algorithm

3 steps

1Step 1: Initialize an array to hold the exclusive time for each function and a stack to track active function calls.
2Step 2: Iterate through each log entry, updating the current time based on whether the log indicates a 'start' or 'end'.
3Step 3: When a function ends, calculate the time spent and update the exclusive time for the function at the top of the stack.

solution.py18 lines

1# Full working Python code
2
3def exclusiveTime(n, logs):
4    exclusive_time = [0] * n
5    stack = []
6    prev_time = 0
7    for log in logs:
8        func_id, status, time = log.split(':')
9        func_id, time = int(func_id), int(time)
10        if status == 'start':
11            if stack:
12                exclusive_time[stack[-1]] += time - prev_time
13            stack.append(func_id)
14            prev_time = time
15        else:
16            exclusive_time[stack.pop()] += time - prev_time + 1
17            prev_time = time + 1
18    return exclusive_time

ℹ

Complexity note: This complexity is linear because we process each log entry exactly once. The stack operations (push and pop) are also efficient, keeping the overall time complexity to O(n).

1Understanding the stack structure is crucial for managing nested function calls.
2Accurate time tracking is essential to calculate exclusive time correctly.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.