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Java-based LeetCode algorithm problem solutions, regularly updated

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2715\. Timeout Cancellation

Easy

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`