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2666. Allow One Function Call

Description

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

Solutions

  • function once<T extends (...args: any[]) => any>(
        fn: T,
    ): (...args: Parameters<T>) => ReturnType<T> | undefined {
        let called = false;
        return function (...args) {
            if (!called) {
                called = true;
                return fn(...args);
            }
        };
    }
    
    /**
     * let fn = (a,b,c) => (a + b + c)
     * let onceFn = once(fn)
     *
     * onceFn(1,2,3); // 6
     * onceFn(2,3,6); // returns undefined without calling fn
     */
    
    

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