916 Word Subsets 單詞子集
Description:
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.
For example, "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:
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"]
Example 3:
Input: words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["e","oo"]
Output: ["facebook","google"]
Example 4:
Input: words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["lo","eo"]
Output: ["google","leetcode"]
Example 5:
Input: words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["ec","oc","ceo"]
Output: ["facebook","leetcode"]
Constraints:
1 <= words1.length, words2.length <= 10^4
1 <= words1[i].length, words2[i].length <= 10
words1[i] and words2[i] consist only of lowercase English letters.
All the strings of words1 are unique.
題目描述:
我們給出兩個單詞數組 A 和 B。每個單詞都是一串小寫字母。
現在,如果 b 中的每個字母都出現在 a 中,包括重複出現的字母,那麼稱單詞 b 是單詞 a 的子集。 例如,“wrr” 是 “warrior” 的子集,但不是 “world” 的子集。
如果對 B 中的每一個單詞 b,b 都是 a 的子集,那麼我們稱 A 中的單詞 a 是通用的。
你可以按任意順序以列表形式返回 A 中所有的通用單詞。
示例 :
示例 1:
輸入:A = ["amazon","apple","facebook","google","leetcode"], B = ["e","o"]
輸出:["facebook","google","leetcode"]
示例 2:
輸入:A = ["amazon","apple","facebook","google","leetcode"], B = ["l","e"]
輸出:["apple","google","leetcode"]
示例 3:
輸入:A = ["amazon","apple","facebook","google","leetcode"], B = ["e","oo"]
輸出:["facebook","google"]
示例 4:
輸入:A = ["amazon","apple","facebook","google","leetcode"], B = ["lo","eo"]
輸出:["google","leetcode"]
示例 5:
輸入:A = ["amazon","apple","facebook","google","leetcode"], B = ["ec","oc","ceo"]
輸出:["facebook","leetcode"]
提示:
1 <= A.length, B.length <= 10000
1 <= A[i].length, B[i].length <= 10
A[i] 和 B[i] 只由小寫字母組成。
A[i] 中所有的單詞都是獨一無二的,也就是說不存在 i != j 使得 A[i] == A[j]。
思路:
計數
先記錄 B 中出現的所有字符串的所有字符的最大出現頻率
然後遍歷 A 中所有字符串, 如果 A 中的字符串中字符頻率不小於 B 的最大出現頻率則加入 result
時間複雜度爲 O(m + n), 空間複雜度爲 O(m + n), m, n 分別爲 A, B 數組的長度
代碼:
C++:
class Solution
{
public:
vector<string> wordSubsets(vector<string>& words1, vector<string>& words2)
{
vector<int> count(26, 0);
for (const auto& word : words2)
{
vector<int> cur(26, 0);
for (const auto& c : word) ++cur[c - 'a'];
for (int i = 0; i < 26; i++) count[i] = max(count[i], cur[i]);
}
vector<string> result;
for (const auto& word : words1)
{
vector<int> cur(26, 0);
for (const auto& c : word) ++cur[c - 'a'];
bool is_found = true;
for (int i = 0; i < 26; i++)
{
if (count[i] > cur[i])
{
is_found = false;
break;
}
}
if (is_found) result.emplace_back(word);
}
return result;
}
};
Java:
class Solution {
public List<String> wordSubsets(String[] words1, String[] words2) {
int count[] = new int[26];
for (String word : words2) {
int cur[] = new int[26];
for (char c : word.toCharArray()) ++cur[c - 'a'];
for (int i = 0; i < 26; i++) count[i] = Math.max(count[i], cur[i]);
}
List<String> result = new ArrayList<>();
for (String word : words1) {
int cur[] = new int[26];
for (char c : word.toCharArray()) ++cur[c - 'a'];
boolean isFound = true;
for (int i = 0; i < 26; i++)
{
if (count[i] > cur[i])
{
isFound = false;
break;
}
}
if (isFound) result.add(word);
}
return result;
}
}
Python:
class Solution:
def wordSubsets(self, words1: List[str], words2: List[str]) -> List[str]:
c2 = reduce(collections.Counter.__or__, map(Counter, words2))
return list(filter(lambda word: all(v <= Counter(word)[k] for k, v in c2.items()), words1))