- Difficulty: Easy
- Tags: LeetCode, Easy, leetcode-2715
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 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, before the delay of t
milliseconds, the function cancelFn
is invoked, it should cancel the delayed execution of fn
. Otherwise, if cancelFn
is not invoked within the specified delay t
, fn
should be executed with the provided args
as arguments.
Example 1:
Input: fn = (x) => x * 5, args = [2], t = 20 Output: [{"time": 20, "returned": 10}] Explanation: const cancelTimeMs = 50; const cancelFn = cancellable((x) => x * 5, [2], 20); setTimeout(cancelFn, cancelTimeMs); The cancellation was scheduled to occur after a delay of cancelTimeMs (50ms), which happened after the execution of fn(2) at 20ms.
Example 2:
Input: fn = (x) => x**2, args = [2], t = 100 Output: [] Explanation: const cancelTimeMs = 50; const cancelFn = cancellable((x) => x**2, [2], 100); setTimeout(cancelFn, cancelTimeMs); The cancellation was scheduled to occur after a delay of cancelTimeMs (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 Output: [{"time": 30, "returned": 8}] Explanation: const cancelTimeMs = 100; const cancelFn = cancellable((x1, x2) => x1 * x2, [2,4], 30); setTimeout(cancelFn, cancelTimeMs); The cancellation was scheduled to occur after a delay of cancelTimeMs (100ms), which happened after the execution of fn(2,4) at 30ms.
Constraints:
fn
is a functionargs
is a valid JSON array1 <= args.length <= 10
20 <= t <= 1000
10 <= cancelTimeMs <= 1000