[LeetCode] 1935. Maximum Number of Words You Can Type

There is a malfunctioning keyboard where some letter keys do not work. All other keys on the keyboard work properly.

Given a string text of words separated by a single space (no leading or trailing spaces) and a string brokenLetters of all distinct letter keys that are broken, return the number of words in text you can fully type using this keyboard.

Example 1:
Input: text = "hello world", brokenLetters = "ad"
Output: 1
Explanation: We cannot type "world" because the 'd' key is broken.

Example 2:
Input: text = "leet code", brokenLetters = "lt"
Output: 1
Explanation: We cannot type "leet" because the 'l' and 't' keys are broken.

Example 3:
Input: text = "leet code", brokenLetters = "e"
Output: 0
Explanation: We cannot type either word because the 'e' key is broken.

Constraints:
1 <= text.length <= 104
0 <= brokenLetters.length <= 26
text consists of words separated by a single space without any leading or trailing spaces.
Each word only consists of lowercase English letters.
brokenLetters consists of distinct lowercase English letters.

可以输入的最大单词数。

键盘出现了一些故障，有些字母键无法正常工作。而键盘上所有其他键都能够正常工作。

给你一个由若干单词组成的字符串 text ，单词间由单个空格组成（不含前导和尾随空格）；另有一个字符串 brokenLetters ，由所有已损坏的不同字母键组成，返回你可以使用此键盘完全输入的 text 中单词的数目。

思路

思路是 hashmap。先将字符串 text 按空格分割成若干个单词，然后判断每个单词中是否包含 brokenLetters 里所有的单词。

复杂度

时间O(mn)
空间O(1)

代码

Java实现

class Solution {
    public int canBeTypedWords(String text, String brokenLetters) {
        int count = 0;
        String[] words = text.split(" ");
        for (String word : words) {
            boolean flag = false;
            for (char c : brokenLetters.toCharArray()) {
                if (word.indexOf(c) != -1) {
                    flag = true;
                    count++;
                    break;
                }
            }
        }
        return words.length - count;
    }
}