problemeasyalgorithmsleetcode-2725leetcode 2725leetcode2725

Interval Cancellation

EasyUpdated: Aug 2, 2025
Practice on:
LeetCode

Problem

Given a function fn, an array of arguments args, and an interval time t, return a cancel function cancelFn.

After a delay of cancelTimeMs, the returned cancel function cancelFn will be invoked.

setTimeout(cancelFn, cancelTimeMs)

The function fn should be called with args immediately and then called again every t milliseconds until cancelFn is called at cancelTimeMs ms.

Examples

Example 1

Input: fn = (x) => x * 2, args = [4], t = 35
Output: 
[
   {"time": 0, "returned": 8},
   {"time": 35, "returned": 8},
   {"time": 70, "returned": 8},
   {"time": 105, "returned": 8},
   {"time": 140, "returned": 8},
   {"time": 175, "returned": 8}
]
Explanation: 
const cancelTimeMs = 190;
const cancelFn = cancellable((x) => x * 2, [4], 35);
setTimeout(cancelFn, cancelTimeMs);

Every 35ms, fn(4) is called. Until t=190ms, then it is cancelled.
1st fn call is at 0ms. fn(4) returns 8.
2nd fn call is at 35ms. fn(4) returns 8.
3rd fn call is at 70ms. fn(4) returns 8.
4th fn call is at 105ms. fn(4) returns 8.
5th fn call is at 140ms. fn(4) returns 8.
6th fn call is at 175ms. fn(4) returns 8.
Cancelled at 190ms

Example 2

Input: fn = (x1, x2) => (x1 * x2), args = [2, 5], t = 30
Output: 
[
   {"time": 0, "returned": 10},
   {"time": 30, "returned": 10},
   {"time": 60, "returned": 10},
   {"time": 90, "returned": 10},
   {"time": 120, "returned": 10},
   {"time": 150, "returned": 10}
]
Explanation: 
const cancelTimeMs = 165; 
const cancelFn = cancellable((x1, x2) => (x1 * x2), [2, 5], 30) 
setTimeout(cancelFn, cancelTimeMs)

Every 30ms, fn(2, 5) is called. Until t=165ms, then it is cancelled.
1st fn call is at 0ms 
2nd fn call is at 30ms 
3rd fn call is at 60ms 
4th fn call is at 90ms 
5th fn call is at 120ms 
6th fn call is at 150ms
Cancelled at 165ms

Example 3

Input: fn = (x1, x2, x3) => (x1 + x2 + x3), args = [5, 1, 3], t = 50
Output: 
[
   {"time": 0, "returned": 9},
   {"time": 50, "returned": 9},
   {"time": 100, "returned": 9},
   {"time": 150, "returned": 9}
]
Explanation: 
const cancelTimeMs = 180;
const cancelFn = cancellable((x1, x2, x3) => (x1 + x2 + x3), [5, 1, 3], 50)
setTimeout(cancelFn, cancelTimeMs)

Every 50ms, fn(5, 1, 3) is called. Until t=180ms, then it is cancelled. 
1st fn call is at 0ms
2nd fn call is at 50ms
3rd fn call is at 100ms
4th fn call is at 150ms
Cancelled at 180ms

Constraints

  • fn is a function
  • args is a valid JSON array
  • 1 <= args.length <= 10
  • 30 <= t <= 100
  • 10 <= cancelTimeMs <= 500

Solution

Method 1 – setInterval with Cancel Function

Intuition

We want to repeatedly call a function at a fixed interval, and be able to cancel the repeated calls at any time. We use setInterval to schedule repeated calls and return a cancel function that clears the interval.

Approach

  1. Call fn(...args) immediately.
  2. Use setInterval to call fn(...args) every t milliseconds.
  3. Return a cancel function that, when called, clears the interval using clearInterval.

Code

JavaScript
function cancellable(fn, args, t) {
  fn(...args);
  const id = setInterval(() => fn(...args), t);
  return function cancelFn() {
    clearInterval(id);
  };
}
TypeScript
function cancellable(fn: (...args: any[]) => any, args: any[], t: number): () => void {
  fn(...args);
  const id = setInterval(() => fn(...args), t);
  return function cancelFn() {
    clearInterval(id);
  };
}

Complexity

  • ⏰ Time complexity: O(1) — Scheduling and cancelling intervals are constant time operations.
  • 🧺 Space complexity: O(1) — Only a single interval id is stored.

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