1684. Count the Number of Consistent Strings

# 1684. Count the Number of Consistent Strings#

## 题目 #

You are given a string `allowed` consisting of distinct characters and an array of strings `words`. A string is consistent if all characters in the string appear in the string `allowed`.

Return the number of consistent strings in the array `words`.

Example 1:

``````Input: allowed = "ab", words = ["ad","bd","aaab","baa","badab"]
Output: 2
Explanation: Strings "aaab" and "baa" are consistent since they only contain characters 'a' and 'b'.

``````

Example 2:

``````Input: allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"]
Output: 7
Explanation: All strings are consistent.

``````

Example 3:

``````Input: allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"]
Output: 4
Explanation: Strings "cc", "acd", "ac", and "d" are consistent.

``````

Constraints:

• `1 <= words.length <= 104`
• `1 <= allowed.length <= 26`
• `1 <= words[i].length <= 10`
• The characters in `allowed` are distinct.
• `words[i]` and `allowed` contain only lowercase English letters.

## 解题思路 #

• 简单题。先将 `allowed` 转化成 map。将 `words` 数组中每个单词的字符都在 map 中查找一遍，如果都存在就累加 res。如果有不存在的字母，不累加。最终输出 res 即可。

## 代码 #

``````package leetcode

func countConsistentStrings(allowed string, words []string) int {
allowedMap, res, flag := map[rune]int{}, 0, true
for _, str := range allowed {
allowedMap[str]++
}
for i := 0; i < len(words); i++ {
flag = true
for j := 0; j < len(words[i]); j++ {
if _, ok := allowedMap[rune(words[i][j])]; !ok {
flag = false
break
}
}
if flag {
res++
}
}
return res
}
``````

Apr 8, 2023