2715. Timeout Cancellation

Problem

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

After a delay of t, fn should be called with args passed as parameters unless cancelFn was invoked before the delay of t milliseconds elapses, specifically at cancelT ms. In that case, fn should never be called.

Example 1:

Input: fn = (x) => x * 5, args = [2], t = 20, cancelT = 50
Output: [{"time": 20, "returned": 10}]
Explanation:
const cancel = cancellable((x) => x * 5, [2], 20); // fn(2) called at t=20ms
setTimeout(cancel, 50);

The cancellation was scheduled to occur after a delay of cancelT (50ms), which happened after the execution of fn(2) at 20ms.

Example 2:

Input: fn = (x) => x**2, args = [2], t = 100, cancelT = 50
Output: []
Explanation:
const cancel = cancellable((x) => x**2, [2], 100); // fn(2) not called
setTimeout(cancel, 50);

The cancellation was scheduled to occur after a delay of cancelT (50ms), which happened before the execution of fn(2) at 100ms, resulting in fn(2) never being called.

Example 3:

Input: fn = (x1, x2) => x1 * x2, args = [2,4], t = 30, cancelT = 100
Output: [{"time": 30, "returned": 8}]
Explanation:
const cancel = cancellable((x1, x2) => x1 * x2, [2,4], 30); // fn(2,4) called at t=30ms
setTimeout(cancel, 100);

The cancellation was scheduled to occur after a delay of cancelT (100ms), which happened after the execution of fn(2,4) at 30ms.

Constraints:

• fn is a function
• args is a valid JSON array
• 1 <= args.length <= 10
• 20 <= t <= 1000
• 10 <= cancelT <= 1000

Solution (Javascript)

/**
 * @param {Function} fn
 * @param {Array} args
 * @param {number} t
 * @return {Function}
 */
var cancellable = function (fn, args, t) {
  // call setTimeout, which is set to call fn after t amount of time
  var timeout = setTimeout(() => fn(...args), t);

  // define a clearTimeout
  var cancelFn = () => clearTimeout(timeout);

  // When/if we call the function, it will return cancelFn,
  // and since the return line calls (and consequentially executes)
  // cancelFn, timeout will be cancelled, thereby cancelling fn
  return cancelFn;
};

/**
 *  const result = []
 *
 *  const fn = (x) => x * 5
 *  const args = [2], t = 20, cancelT = 50
 *
 *  const start = performance.now()
 *
 *  const log = (...argsArr) => {
 *      const diff = Math.floor(performance.now() - start);
 *      result.push({"time": diff, "returned": fn(...argsArr))
 *  }
 *
 *  const cancel = cancellable(log, args, t);
 *
 *  const maxT = Math.max(t, cancelT)
 *
 *  setTimeout(() => {
 *     cancel()
 *  }, cancelT)
 *
 *  setTimeout(() => {
 *     console.log(result) // [{"time":20,"returned":10}]
 *  }, maxT + 15)
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).