LeetCode 2630: Memoize II Solution

Problem Explanation

Explanation

To memoize a function, we can use a hashmap to store the results of previous function calls based on the input arguments. We will convert the input arguments into a unique key to use as the hashmap key. If the key is already present in the hashmap, we retrieve the result from the cache; otherwise, we call the original function and store the result in the cache.

Algorithm:

Create a hashmap to store the cached results.
Define a wrapper function that takes the input arguments, converts them to a unique key, and checks if the key exists in the cache. If it does, return the cached result; otherwise, call the original function, store the result in the cache, and return it.
Return the wrapper function as the memoized version of the original function.

Time Complexity:

Calling the memoized function has a time complexity of O(1) if the result is found in the cache, or O(n) if the result needs to be computed.
Overall time complexity depends on the number of unique inputs and the number of function calls.

Space Complexity:

The space complexity is O(n) where n is the number of unique inputs to the function.

Solution Code

import java.util.HashMap;
import java.util.Map;
import java.util.function.BiFunction;

class Memoize {
    public static <T, R> BiFunction<T, T, R> memoize(BiFunction<T, T, R> fn) {
        Map<String, R> cache = new HashMap<>();
        return (a, b) -> {
            String key = a.toString() + b.toString();
            if (cache.containsKey(key)) {
                return cache.get(key);
            } else {
                R result = fn.apply(a, b);
                cache.put(key, result);
                return result;
            }
        };
    }
}

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2630. Memoize II