How do you solve Interval Cancellation on LeetCode?

Interval Cancellation (LeetCode #2725) 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 how setTimeout and setInterval work is crucial for timing functions.

What is the brute force solution for Interval Cancellation?

The brute force approach for Interval Cancellation is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves calling the function immediately and then using a loop with setTimeout to repeatedly call it at the specified interval until the cancellation time is reached. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Interval Cancellation?

Always clarify the timing requirements before coding. Discuss edge cases, such as what happens if cancelTimeMs is less than t. Practice writing asynchronous code to become comfortable with timing functions.

What are common mistakes when solving Interval Cancellation?

Not clearing intervals properly, leading to memory leaks. Confusing the timing of setTimeout and setInterval.

#2725

Interval Cancellation

Easy

Timer FunctionsAsynchronous Programming

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 uses setInterval to repeatedly call the function at the specified interval until the cancellation time is reached. This approach is efficient and leverages JavaScript's built-in timing functions.

⚙️

Algorithm

3 steps

1Step 1: Call the function fn with the provided args immediately and store the result.
2Step 2: Use setInterval to call fn every t milliseconds.
3Step 3: Use setTimeout to clear the interval after cancelTimeMs.

solution.py13 lines

1import time
2
3def cancellable(fn, args, t, cancelTimeMs):
4    results = []
5    results.append({'time': 0, 'returned': fn(*args)})
6    def call_fn():
7        results.append({'time': int(time.time() * 1000), 'returned': fn(*args)})
8    timer = time.time()
9    while time.time() < timer + cancelTimeMs / 1000:
10        call_fn()
11        time.sleep(t / 1000)
12    return results
13

ℹ

Complexity note: The time complexity is linear because we are calling the function a fixed number of times based on the interval and cancellation time.

1Understanding how setTimeout and setInterval work is crucial for timing functions.
2Recognizing the importance of managing intervals and cancellations in asynchronous programming.

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