1178. Number of Valid Words for Each Puzzle

Topic Description

　　This last question is the subject of September 1, 2019 LeetCode week race

Title 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" (doesn't include "a") and "based" (includes "s" which isn't in the puzzle). Return an array answer, where answer[i] is the number of words in the given word list words that are valid with respect to the puzzle puzzles[i].

Sample

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're no valid words for "gaswxyz" cause none of the words in the list contains letter 'g'.

Conditions

1 <= words.length <= 10^5 4 <= words[i].length <= 50 1 <= puzzles.length <= 10^4 puzzles[i].length == 7 words[i][j], puzzles[i][j] are English lowercase letters. Each puzzles[i] doesn't contain repeated characters.

Problem-solving ideas

① first traversal words again, and then the inside of the letters of each word as an index (i.e., inside the word letter -'a 'as an index), the word placed in the corresponding array ② calculating a unique value corresponding to a word 　　I'm using 26 binary representation, for example, expressed as aba 11 　　<< 1 ((corresponding to the letter will not be repeated) - 'a') calculating the position of a character corresponding to 　　The first 11 and the second representation b expressed as a, repeated letters are not considered ③ array traversal puzzles, also calculates a corresponding unique value corresponding to a word, and then using the first letter of a word lookup meet meaning of the title (title requirement is intended, the first letter of a word puzzle which is contained in the word), then traverse word corresponding to the array index ④ determining the subject word meets a condition 　　The method used is the unique value is a unique value determined by the unique word Puzzle value corresponding to the operation of the bit operation and word operation is determined to be equal, the condition is satisfied 　　　　如 　　word ="a"  , puzzle  = "ab" 　　　　The only value of 111 　　　　After the operation is a 1 & 　　It should be noted that: a lower priority than the & operator ==

Code

    vector<int>w;
    vector<int>wf[26];
    bool vis[26];
    vector<int> findNumOfValidWords(vector<string>& words, vector<string>& puzzles) {
        
        for(int i=0;i<words.size();i++){
            memset(vis,false,sizeof(vis));
            int tmp = 0;
            
            for(int j=0;j<words[i].size();j++){
                if(!vis[words[i][j]-'a']){
                    tmp+=(1<<(words[i][j]-'a'));
                    vis[words[i][j]-'a']=true;
                    wf[words[i][j]-'a'].push_back(i);
                    
                }
            }

            w.push_back(tmp);
        }
        
        vector<int>res;
        for(int i=0;i<puzzles.size();i++){
            int index = puzzles[i][0]-'a';
            int h  = 0;
            for(int j=0;j<7;j++){
                h+=(1<<(puzzles[i][j]-'a'));
            }
            int cnt =0 ;
            for(int j=0;j<wf[index].size();j++){        
                if((w[wf[index][j]]&h)==w[wf[index][j]]){
                    cnt++;
                }
            }
            res.push_back(cnt);
        }
        return res;
        
    }

Code