[Funtions, Timeout] Execute Cancellable Function With Delay

https://leetcode.com/problems/execute-cancellable-function-with-delay/description/

Execute Cancellable Function With Delay - LeetCode

Given a function fn, an array or 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 called first. In that case, fn should never be called.

 

 

Example 1:

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

the cancelTime (50ms) is after the delay time (20ms), so fn(2) should be called at t=20ms. The value returned from fn is 10.

Example 2:

Input: fn = (x) => x**2, args = [2], t = 100, cancelTime = 50
Output: []
Explanation: fn(2) was never called because cancelTime (50ms) is before the delay time (100ms).

Example 3:

Input: fn = (x1, x2) => x1 * x2, args = [2,4], t = 30, cancelTime = 100
Output: [{"time": 30, "returned": 8}]
Explanation: fn(2, 4) was called at t=30ms because cancelTime > t.

Constraints:

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

 

solution

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

/**
 *  const result = []
 *
 *  const fn = (x) => x * 5
 *  const args = [2], t = 20, cancelT = 50
 *
 *  const log = (...argsArr) => {
 *      result.push(fn(...argsArr))
 *  }
 *       
 *  const cancel = cancellable(fn, args, t);
 *           
 *  setTimeout(() => {
 *     cancel()
 *     console.log(result) // [{"time":20,"returned":10}]
 *  }, cancelT)
 */