puzzle
string, a word
is valid if both the following conditions are satisfied:word
contains the first letter ofpuzzle
.- For each letter in
word
, that letter is inpuzzle
.
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).
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]
.
Example :
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'.
Constraints:
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.
A:
首先的想法是,把所有的都变成int.
class Solution { public: vector<int> findNumOfValidWords(vector<string>& words, vector<string>& puzzles) { int np = puzzles.size(); vector<int> P(np,0);
for(int i =0;i<np;i++){ // change each puzzle into int for(auto ch : puzzles[i]) P[i] |= (1<<(ch-'a')); } int nw = words.size(); vector<int> W(nw,0); for(int i =0;i<nw;i++){ // change each word into int for(auto ch : words[i]) W[i] |= (1<<(ch-'a')); } vector<int> res(np,0); for(int i=0;i<np;i++){ int first = 1<< (puzzles[i][0]-'a'); for(auto iW : W){ if( ( first & iW) and ( (iW & P[i]) == iW )) res[i]++; } } return res; } };
然而,还是LTE了
好啦,直接看答案了。 key point是 , query的时候,因为puzzles[i].length == 7,所以我们可以直接看检查所有的subset . 这样,最多有2^7==128个
class Solution { public: vector<int> findNumOfValidWords(vector<string>& words, vector<string>& puzzles) { int np = puzzles.size(); vector<int> P(np,0); for(int i =0;i<np;i++){ // change each puzzle into int for(auto ch : puzzles[i]) P[i] |= (1<<(ch-'a')); } int nw = words.size(); unordered_map<int,int> wordCount; for(int i =0;i<nw;i++){ // change each word into int int tmp = 0; for(auto ch : words[i]) tmp |= (1<<(ch-'a')); wordCount[tmp]++; } vector<int> res; for(int i=0;i<np;i++){ int first = 1<< (puzzles[i][0]-'a'); int puz = P[i] - first; vector<int> subset = getAllSubset(puz);// try all permutaion of puz (2^6 = 64 choices) int total = 0; for(int k : subset){//subset is at most 64 long auto iter = wordCount.find(k + first); //find() will not create default object if not presented if(iter != wordCount.end()) total += iter->second; } res.push_back(total); } return res; } private: vector<int> getAllSubset(int val){ // only places of 1, can be 0, no 0 bit change to 1 if(val ==0){ return vector<int>{0}; } int rightMostBitOf1 = val ^(val & (val-1)); auto sub = getAllSubset(val - rightMostBitOf1); int len = sub.size(); for(int i = 0;i<len;i++){ sub.push_back(sub[i]+rightMostBitOf1); } return sub; } };
这里,原版的厉害处是,没有单独计算getAllSubset(), 而是直接用数学方法计算的。
但是感觉会有多余的值。也没搞明白他到底是怎么找到所有的subset的。 不管了。 就这样了
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