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"]
public List<String> wordSubsets(String[] A, String[] B) {
int[] bmax = frequency("");
for (String s : B) {
int[] count = frequency(s);
for (int i = 0; i < 26; i++) {
bmax[i] = Math.max(bmax[i], count[i]);
}
}
List<String> res = new ArrayList<>();
for (String s : A) {
int[] count = frequency(s);
boolean found = true;
for (int i = 0; i < 26; i++) {
if (bmax[i] > count[i]) {
found = false;
break;
}
}
if (found)
res.add(s);
}
return res;
}private helper method
private int[] frequency(String s) {
int[] count = new int[26];
for (char c : s.toCharArray()) {
count[c - 'a']++;
}
return count;
}Compile with javac Solution.java and run with java Solution.
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