916. Word Subsets

We are given two arrays A and B of words.  Each word is a string of lowercase letters.

Now, say that word b is a subset of word a if every letter in b occurs in a, including multiplicity.  For example, "wrr" is a subset of "warrior", but is not a subset of "world".

Now say a word a from A is  universal  if for every b in Bb is a subset of a.

Return a list of all universal words in A.  You can return the words in any order.

Example 1:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","o"]
Output: ["facebook","google","leetcode"]

Example 2:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["l","e"]
Output: ["apple","google","leetcode"]

Example 3:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","oo"]
Output: ["facebook","google"]

Example 4:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["lo","eo"]
Output: ["google","leetcode"]

Example 5:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["ec","oc","ceo"]
Output: ["facebook","leetcode"]

Note:

  1. 1 <= A.length, B.length <= 10000

  2. 1 <= A[i].length, B[i].length <= 10

  3. A[i] and B[i] consist only of lowercase letters.

  4. All words in A[i] are unique: there isn’t i != j with A[i] == A[j].

    这道题定义了两个单词之间的一种子集合关系,就是说假如单词b中的每个字母都在单词a中出现了(包括重复字母),就说单词b是单词a的子集合。现在给了两个单词集合A和B,让找出集合A中的所有满足要求的单词,使得集合B中的所有单词都是其子集合。配合上题目中给的一堆例子,意思并不难理解,根据子集合的定义关系,其实就是说若单词a中的每个字母的出现次数都大于等于单词b中每个字母的出现次数,单词b就一定是a的子集合。现在由于集合B中的所有单词都必须是A中某个单词的子集合,那么其实只要对于每个字母,都统计出集合B中某个单词中出现的最大次数,比如对于这个例子,B=[“eo”,”oo”],其中e最多出现1次,而o最多出现2次,那么只要集合A中有单词的e出现不少1次,o出现不少于2次,则集合B中的所有单词一定都是其子集合。这就是本题的解题思路,这里使用一个大小为 26 的一维数组 charCnt 来统计集合B中每个字母的最大出现次数,而将统计每个单词的字母次数的操作放到一个子函数 helper 中,当 charCnt 数组更新完毕后,下面就开始检验集合A中的所有单词了。对于每个遍历到的单词,还是要先统计其每个字母的出现次数,然后跟 charCnt 中每个位置上的数字比较,只要均大于等于 charCnt 中的数字,就可以加入到结果 res 中了,参见代码如下:

    class Solution {
    public:
    vector wordSubsets(vector& A, vector& B) {

       vector<string> res;
       vector<int> charCnt(26);
       for (string &b : B) {
           vector<int> t = helper(b);
           for (int i = 0; i < 26; ++i) {
               charCnt[i] = max(charCnt[i], t[i]);
           }
       }
       for (string &a : A) {
           vector<int> t = helper(a);
           int i = 0;
           for (; i < 26; ++i) {
               if (t[i] < charCnt[i]) break;
           }
           if (i == 26) res.push_back(a);
       }
       return res;
    

    }
    vector helper(string& word) {

       vector<int> res(26);
       for (char c : word) ++res[c - 'a'];
       return res;
    

    }
    };

    Github 同步地址:

https://github.com/grandyang/leetcode/issues/916

参考资料:

https://leetcode.com/problems/word-subsets/

https://leetcode.com/problems/word-subsets/discuss/175854/C%2B%2BJavaPython-Straight-Forward

LeetCode All in One 题目讲解汇总(持续更新中…)


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