README.md

1178. Number of Valid Words for Each Puzzle

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 (Rust)

1. Solution

use std::collections::HashMap;

impl Solution {
    pub fn find_num_of_valid_words(words: Vec<String>, puzzles: Vec<String>) -> Vec<i32> {
        let mut count = HashMap::new();
        let mut answer = vec![0; puzzles.len()];

        for word in words {
            let bits = word.bytes().fold(0_i32, |b, c| b | (1 << (c - b'a')));

            if bits.count_ones() < 8 {
                *count.entry(bits).or_insert(0) += 1;
            }
        }

        for i in 0..answer.len() {
            let puzzle = puzzles[i].as_bytes();
            let mut combins = vec![1 << (puzzle[0] - b'a')];
            answer[i] += count.get(&combins[0]).unwrap_or(&0);

            for j in 1..puzzle.len() {
                for k in 0..combins.len() {
                    let combin = combins[k] | (1 << (puzzle[j] - b'a'));
                    combins.push(combin);
                    answer[i] += count.get(&combin).unwrap_or(&0);
                }
            }
        }

        answer
    }
}