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Java-based LeetCode algorithm problem solutions, regularly updated

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2666\. Allow One Function Call

Easy

Given a function `fn`, return a new function that is identical to the original function except that it ensures `fn` is called at most once.

*   The first time the returned function is called, it should return the same result as `fn`.
*   Every subsequent time it is called, it should return `undefined`.

**Example 1:**

**Input:** fn = (a,b,c) => (a + b + c), calls = [[1,2,3],[2,3,6]]

**Output:** [{"calls":1,"value":6}]

**Explanation:** 

    const onceFn = once(fn); 
    onceFn(1, 2, 3); // 6 
    onceFn(2, 3, 6); // undefined, fn was not called

**Example 2:**

**Input:** fn = (a,b,c) => (a \* b \* c), calls = [[5,7,4],[2,3,6],[4,6,8]]

**Output:** [{"calls":1,"value":140}]

**Explanation:** 

    const onceFn = once(fn); 
    onceFn(5, 7, 4); // 140 
    onceFn(2, 3, 6); // undefined, fn was not called 
    onceFn(4, 6, 8); // undefined, fn was not called

**Constraints:**

*   `1 <= calls.length <= 10`
*   `1 <= calls[i].length <= 100`
*   `2 <= JSON.stringify(calls).length <= 1000`