You are given two string arrays words1 and words2.
A string b is a subset of string a if every letter in b occurs in a including multiplicity.
"wrr" is a subset of "warrior" but is not a subset of "world".A string a from words1 is universal if for every string b in words2, b is a subset of a.
Return an array of all the universal strings in words1. You may return the answer in any order.
Example 1:
Input: words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["e","o"] Output: ["facebook","google","leetcode"]
Example 2:
Input: words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["l","e"] Output: ["apple","google","leetcode"]
Constraints:
1 <= words1.length, words2.length <= 1041 <= words1[i].length, words2[i].length <= 10words1[i] and words2[i] consist only of lowercase English letters.words1 are unique.class Solution {
public:
vector<string> wordSubsets(vector<string>& A, vector<string>& B) {
vector<int>freq(26,0);
for(auto x:B){
vector<int>temp(26,0);
for(auto y:x){
temp[y-'a']++;
freq[y-'a'] = max(freq[y-'a'],temp[y-'a']);
}
}
vector<string>res;
for(auto x:A){
vector<int>temp(26,0);
for(auto y:x)
temp[y-'a']++;
bool flag=true;
for(int i=0 ; i<26 ; i++)
if(freq[i]>temp[i]) {
flag=false;
break;
}
if(flag) res.push_back(x);
}
return res;
}
};class Solution {
public List<String> wordSubsets(String[] words1, String[] words2) {
int[] universalFreq = new int[26];
for (String s : words2) {
int[] freq = new int[26];
for (char c : s.toCharArray())
freq[c - 'a']++;
for (int i = 0; i < 26; i++)
universalFreq[i] = Math.max(universalFreq[i], freq[i]);
}
List<String> universalStrings = new ArrayList<>();
for (String s : words1) {
int[] freq = new int[26];
for (char c : s.toCharArray())
freq[c - 'a']++;
int i;
for (i = 0; i < 26 && universalFreq[i] <= freq[i]; i++)
;
if (i == 26)
universalStrings.add(s);
}
return universalStrings;
}
}