LeetCode 2666: Allow One Function Call Solution

Problem Explanation

Explanation:

To solve this problem, we need to create a new function that wraps the original function fn. This new function should ensure that fn is called at most once. We can achieve this by keeping track of whether fn has been called before. If it has been called, we return undefined. Otherwise, we call fn and record the result.

Here is the algorithm:

Create a boolean variable called initialized to false to track if fn has been called.
Create a new function onceFn that takes the same arguments as fn.
Inside onceFn, check if called is false.
- If called is false, call fn with the provided arguments and store the result.
  - Set called to true.
  - Return the result of calling fn.
- If called is true, return undefined.

Time complexity: O(1) for each function call. Space complexity: O(1).

Solution Code

import java.util.*;
import java.util.function.Function;

class Solution {
    public static Function<int[], Integer> once(Function<int[], Integer> fn) {
        boolean called = false;
        int result = 0;

        return (args) -> {
            if (!called) {
                result = fn.apply(args);
                called = true;
                return result;
            } else {
                return null;
            }
        };
    }
}

Frequently Asked Questions

How to solve LeetCode 2666 (Allow One Function Call)?

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What is the time complexity of LeetCode 2666 (Allow One Function Call)?

The time complexity for LeetCode 2666 (Allow One Function Call) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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2666. Allow One Function Call