Formatted question description: https://leetcode.ca/all/916.html
916. Word Subsets (Medium)
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]
andB[i]
consist only of lowercase letters.- All words in
A[i]
are unique: there isn'ti != j
withA[i] == A[j]
.
Related Topics:
String
Solution 1.
// OJ: https://leetcode.com/problems/word-subsets/
// Time: O((A + B) * N) where N is the maximum length of string
// Space: O(1)
class Solution {
public:
vector<string> wordSubsets(vector<string>& A, vector<string>& B) {
int N = A.size(), target[26] = {}, j = 0;
for (auto &b : B) {
int cnt[26] = {};
for (char c : b) cnt[c - 'a']++;
for (int i = 0; i < 26; ++i) target[i] = max(target[i], cnt[i]);
}
for (int i = 0; i < N; ++i) {
int cnt[26] = {};
for (char c : A[i]) cnt[c - 'a']++;
bool valid = true;
for (int k = 0; k < 26 && valid; ++k) valid = target[k] <= cnt[k];
if (valid) A[j++] = A[i];
}
A.resize(j);
return A;
}
};
Java
class Solution {
public List<String> wordSubsets(String[] A, String[] B) {
List<String> universalList = new ArrayList<String>();
int[] counts = getMaxCounts(B);
for (String word : A) {
int[] wordCounts = getWordCounts(word);
boolean flag = true;
for (int i = 0; i < 26; i++) {
if (wordCounts[i] < counts[i]) {
flag = false;
break;
}
}
if (flag)
universalList.add(word);
}
return universalList;
}
public int[] getMaxCounts(String[] array) {
int[] counts = new int[26];
for (String word : array) {
int[] curCounts = new int[26];
char[] wordArray = word.toCharArray();
for (char c : wordArray)
curCounts[c - 'a']++;
for (int i = 0; i < 26; i++)
counts[i] = Math.max(counts[i], curCounts[i]);
}
return counts;
}
public int[] getWordCounts(String word) {
int[] wordCounts = new int[26];
char[] wordArray = word.toCharArray();
for (char c : wordArray)
wordCounts[c - 'a']++;
return wordCounts;
}
}