LeetCode-in-Java

1178. Number of Valid Words for Each Puzzle

Hard

With respect to a given puzzle string, a word is valid if both the following conditions are satisfied:

• word contains the first letter of puzzle.
• For each letter in word, that letter is in puzzle.
  • For example, if the puzzle is "abcdefg", then valid words are "faced", "cabbage", and "baggage", while
  • invalid words are "beefed" (does not include 'a') and "based" (includes 's' which is not in the puzzle).

Return an array answer, where answer[i] is the number of words in the given word list words that is valid with respect to the puzzle puzzles[i].

Example 1:

Input: words = [“aaaa”,”asas”,”able”,”ability”,”actt”,”actor”,”access”], puzzles = [“aboveyz”,”abrodyz”,”abslute”,”absoryz”,”actresz”,”gaswxyz”]

Output: [1,1,3,2,4,0]

Explanation:

1 valid word for “aboveyz” : “aaaa”

1 valid word for “abrodyz” : “aaaa”

3 valid words for “abslute” : “aaaa”, “asas”, “able”

2 valid words for “absoryz” : “aaaa”, “asas”

4 valid words for “actresz” : “aaaa”, “asas”, “actt”, “access”

There are no valid words for “gaswxyz” cause none of the words in the list contains letter ‘g’.

Example 2:

Input: words = [“apple”,”pleas”,”please”], puzzles = [“aelwxyz”,”aelpxyz”,”aelpsxy”,”saelpxy”,”xaelpsy”]

Output: [0,1,3,2,0]

Constraints:

• 1 <= words.length <= 105
• 4 <= words[i].length <= 50
• 1 <= puzzles.length <= 104
• puzzles[i].length == 7
• words[i] and puzzles[i] consist of lowercase English letters.
• Each puzzles[i] does not contain repeated characters.

Solution

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;

public class Solution {
    public List<Integer> findNumOfValidWords(String[] words, String[] puzzles) {
        List<Integer> ans = new ArrayList<>();
        HashMap<Integer, Integer> map = new HashMap<>();
        for (String word : words) {
            int wordmask = createMask(word);
            if (map.containsKey(wordmask)) {
                int oldfreq = map.get(wordmask);
                int newfreq = oldfreq + 1;
                map.put(wordmask, newfreq);
            } else {
                map.put(wordmask, 1);
            }
        }
        for (String puzzle : puzzles) {
            int puzzlemask = createMask(puzzle);
            char firstChar = puzzle.charAt(0);
            int first = 1 << (firstChar - 'a');
            int sub = puzzlemask;
            int count = 0;
            while (sub != 0) {
                boolean firstCharPresent = (sub & first) == first;
                boolean wordvalid = map.containsKey(sub);
                if (firstCharPresent && wordvalid) {
                    count += map.get(sub);
                }

                sub = (sub - 1) & puzzlemask;
            }
            ans.add(count);
        }
        return ans;
    }

    private int createMask(String str) {
        int mask = 0;
        for (int i = 0; i < str.length(); i++) {
            int bit = str.charAt(i) - 'a';
            mask = (mask | (1 << bit));
        }
        return mask;
    }
}

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