Word Subsets
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 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"]
Note:
1 <= A.length, B.length <= 100001 <= A[i].length, B[i].length <= 10A[i]andB[i]consist only of lowercase letters.- All words in
A[i]are unique: there isn’ti != jwithA[i] == A[j].
class Solution {
public:
vector<string> wordSubsets(vector<string>& A, vector<string>& B) {
int bCount[26] = {0};
for (auto word : B) {
int wordCount[26] = {0};
for (auto ch : word)
++wordCount[ch - 'a'];
for (int i = 0; i < 26; ++i)
bCount[i] = max(bCount[i], wordCount[i]);
}
vector<string> universal;
for (auto word : A) {
int count[26] = {0};
for (auto ch : word)
++count[ch - 'a'];
bool isUniversal = true;
for (int i=0; i < 26; ++i) {
if(count[i] < bCount[i]) {
isUniversal = false;
break;
}
}
if (isUniversal)
universal.push_back(word);
}
return universal;
}
};