0916. Word Subsets

916. Word Subsets #

题目 #

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 aincluding 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 Bb 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. 1 <= A.length, B.length <= 10000
  2. 1 <= A[i].length, B[i].length <= 10
  3. A[i] and B[i] consist only of lowercase letters.
  4. All words in A[i] are unique: there isn’t i != j with A[i] == A[j].

题目大意 #

我们给出两个单词数组 A 和 B。每个单词都是一串小写字母。现在,如果 b 中的每个字母都出现在 a 中,包括重复出现的字母,那么称单词 b 是单词 a 的子集。 例如,“wrr” 是 “warrior” 的子集,但不是 “world” 的子集。如果对 B 中的每一个单词 b,b 都是 a 的子集,那么我们称 A 中的单词 a 是通用的。你可以按任意顺序以列表形式返回 A 中所有的通用单词。

解题思路 #

  • 简单题。先统计出 B 数组中单词每个字母的频次,再在 A 数组中依次判断每个单词是否超过了这个频次,如果超过了即输出。

代码 #

package leetcode

func wordSubsets(A []string, B []string) []string {
	var counter [26]int
	for _, b := range B {
		var m [26]int
		for _, c := range b {
			j := c - 'a'
			m[j]++
		}
		for i := 0; i < 26; i++ {
			if m[i] > counter[i] {
				counter[i] = m[i]
			}
		}
	}
	var res []string
	for _, a := range A {
		var m [26]int
		for _, c := range a {
			j := c - 'a'
			m[j]++
		}
		ok := true
		for i := 0; i < 26; i++ {
			if m[i] < counter[i] {
				ok = false
				break
			}
		}
		if ok {
			res = append(res, a)
		}
	}
	return res
}

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