LeetCode 916. Word Subsets

原题链接在这里：https://leetcode.com/problems/word-subsets/

题目：

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 B, b 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 <= A.length, B.length <= 10000
1 <= A[i].length, B[i].length <= 10
A[i] and B[i] consist only of lowercase letters.
All words in A[i] are unique: there isn't i != j with A[i] == A[j].

题解：

String a in A, if it is universal for every b in B, it must cover all the letters and max corresponding multiplicity in B.

Thus construct a map to maintain all letters and max corresponding multiplicity in B.

Then for each A, if it could cover this map, then add it to the res.

Time Complexity: O(m*n + p*q). m = A.length. n = average length of string in A. p = B.length. q = average length of string in B.

Space: O(1).

AC Java:

 class Solution {

     public List<String> wordSubsets(String[] A, String[] B) {

         List<String> res = new ArrayList<>();

         if(A == null || A.length == 0){

             return res;

         }

         if(B == null || B.length == 0){

             return Arrays.asList(A);

         }

         int [] map = new int[26];

         for(String b : B){

             int [] bMap = new int[26];

             for(char c : b.toCharArray()){

                 bMap[c - 'a']++;

             }

             for(int i = 0; i<26; i++){

                 map[i] = Math.max(map[i], bMap[i]);

             }

         }

         for(String a : A){

             int [] aMap = new int[26];

             for(char c : a.toCharArray()){

                 aMap[c-'a']++;

             }

             boolean isNotSubSet = false;

             for(int i = 0; i<26; i++){

                 if(aMap[i] < map[i]){

                     isNotSubSet = true;

                     break;

                 }

             }

             if(!isNotSubSet){

                 res.add(a);

             }

         }

         return res;

     }

 }

巴特西

LeetCode 916. Word Subsets

最新文章

热门文章