[LeetCode] 318. Maximum Product of Word Lengths

 Problem : https://leetcode.com/problems/maximum-product-of-word-lengths/

Calculate hash key base on the letters.  Then wordA and wordB have common letter if keyA & keyB != 0

Time Complexity = O ( N ** 2 )

class Solution {
    public int maxProduct(String[] words) {
        final int[] keys = Arrays.stream(words)
            .mapToInt(this::keyOf)
            .toArray();
        
        return IntStream.range(0, words.length)
            .map(i -> maxProductStartsWithWord(i, words, keys))
            .max()
            .orElse(0);
    }
    
    int maxProductStartsWithWord(int i, String[] words, int[] keys) {
        return IntStream.range(i+1, words.length)
                    .filter(j -> (keys[i] & keys[j]) == 0)
                    .map(j -> words[i].length() * words[j].length())
                    .max()
                    .orElse(0);
    }
    
    int keyOf(String word) {
        return word.chars()
            .reduce(0, (key, w) -> key | (1 << (w - 'a')));
    }
}

Edited on 05/31/2022. Replace with Java functional programming style solution.