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Kotlin-based LeetCode algorithm problem solutions, regularly updated
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`