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].
Solution:
class Solution {
public List<String> wordSubsets(String[] A, String[] B)
{
int[] t = new int[26];
for(String b : B)
{
int[] tmp = new int[26];
for(char c : b.toCharArray()) tmp[c - 'a']++;
for(int i = 0; i < 26; i++) t[i] = Math.max(t[i], tmp[i]);
}
List<String> res = new ArrayList<>();
for(String a : A) if(isUniversal(a, t)) res.add(a);
return res;
}
boolean isUniversal(String s, int[] t)
{
int[] tmp = new int[26];
for(char c : s.toCharArray()) tmp[c - 'a']++;
for(int i = 0; i < 26; i++) if(tmp[i] < t[i]) return false;
return true;
}
}
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