2715. Timeout Cancellation

On this page, we explore LeetCode Problem 2715: Timeout Cancellation. This problem involves delaying the execution of a function and providing a way to cancel it if needed.

📄 Problem Statement

Given a function fn, an array of arguments args, and a timeout t in milliseconds, return a cancel function cancelFn.

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

setTimeout(cancelFn, cancelTimeMs);

Initially, the execution of the function fn should be delayed by t milliseconds. If cancelFn is invoked before t milliseconds, it should cancel the delayed execution of fn. Otherwise, fn should be executed with the provided args.

💡 Example

Input:


const fn = (x) => console.log(x);
const args = [42];
const t = 2000;
const cancelTimeMs = 1000;

const cancelFn = cancellable(fn, args, t);
setTimeout(cancelFn, cancelTimeMs);

Output: undefined (since fn was cancelled before execution)

🧠 Explanation

Use setTimeout to schedule fn(...args) to run after t milliseconds. Capture its timeout ID. Then return a cancelFn function that clears the timeout using clearTimeout. If cancelFn is called before t ms, it will cancel the function from running.

✅ Your Solution


/**
 * @param {Function} fn
 * @param {Array} args
 * @param {number} t
 * @return {Function}
 */
var cancellable = function(fn, args, t) {
    const timeoutId = setTimeout(() => {
        fn(...args);
    }, t);

    return function cancelFn() {
        clearTimeout(timeoutId);
    };
};