2666. Allow One Function Call

Problem

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

Solution (Javascript)

/**
 * @param {Function} fn
 * @return {Function}
 */
var once = function(fn) {

  let hasBeenCalled = false;
  let result;

  return function(...args) {
    if (!hasBeenCalled) {
      result = fn(...args);
      hasBeenCalled = true;
      return result;
    } else {
      return undefined;
    }
  }

};

let fn = (a,b,c) => (a + b + c);
let onceFn = once(fn);

console.log(onceFn(1,2,3)); // 6
console.log(onceFn(2,3,6)); // undefined

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).