LeetCode-in-Java

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

Solution

type Fn = (...args: any[]) => any

function once(fn: Fn): Fn {
    let wasCalled = false
    return function (...args) {
        if (!wasCalled) {
            wasCalled = true
            return fn(...args)
        }
    }
}

export { once }

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