1178. Number of Valid Words for Each Puzzle

Problem Link

Description

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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

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Python3

class Solution:
    def findNumOfValidWords(self, words: List[str], puzzles: List[str]) -> List[int]:
        mp = collections.defaultdict(int)
        res = [0] * len(puzzles)
        
        for word in words:
            mask = 0
            for x in word:
                mask |= (1 << (ord(x) - ord('a')))
            mp[mask] += 1
        
        for i, puzzle in enumerate(puzzles):
            mask = 0
            for x in puzzle:
                mask |= (1 << (ord(x) - ord('a')))
                
            first = 1 << (ord(puzzle[0]) - ord('a'))
            
            sub = mask
            
            while True:
                if (sub & first) == first:
                    res[i] += mp[sub]
                
                if sub == 0: break
                
                sub = (sub - 1) & mask
        
        return res