1178. Number of Valid Words for Each Puzzle

Description

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.

 

Solutions

Solution: Bit Manipulation

• Time complexity: O(words.length + puzzles.length)
• Space complexity: O(words.length)

 

JavaScript

/**
 * @param {string[]} words
 * @param {string[]} puzzles
 * @return {number[]}
 */
const findNumOfValidWords = function (words, puzzles) {
  const BASE_CODE = 'a'.charCodeAt(0);
  const maskMap = new Map();

  const getCode = letter => letter.charCodeAt(0) - BASE_CODE;

  for (const word of words) {
    let mask = 0;

    for (const letter of word) {
      mask |= 1 << getCode(letter);
    }
    maskMap.set(mask, (maskMap.get(mask) ?? 0) + 1);
  }

  return puzzles.map(puzzle => {
    let puzzleMask = 0;

    for (const letter of puzzle) {
      puzzleMask |= 1 << getCode(letter);
    }
    const firstMask = 1 << getCode(puzzle[0]);
    let subMask = puzzleMask;
    let result = 0;

    while (subMask) {
      if ((subMask & firstMask) === firstMask) {
        result += maskMap.get(subMask) ?? 0;
      }
      subMask = (subMask - 1) & puzzleMask;
    }
    return result;
  });
};