LeetCode-in-Java
3136. Valid Word
Easy
A word is considered valid if:
- It contains a minimum of 3 characters.
- It contains only digits (0-9), and English letters (uppercase and lowercase).
- It includes at least one vowel.
- It includes at least one consonant.
You are given a string word.
Return true if word is valid, otherwise, return false.
Notes:
'a','e','i','o','u', and their uppercases are vowels.- A consonant is an English letter that is not a vowel.
Example 1:
Input: word = “234Adas”
Output: true
Explanation:
This word satisfies the conditions.
Example 2:
Input: word = “b3”
Output: false
Explanation:
The length of this word is fewer than 3, and does not have a vowel.
Example 3:
Input: word = “a3$e”
Output: false
Explanation:
This word contains a '$' character and does not have a consonant.
Constraints:
1 <= word.length <= 20wordconsists of English uppercase and lowercase letters, digits,'@','#', and'$'.
Solution
public class Solution {
public boolean isValid(String word) {
if (word.length() < 3) {
return false;
}
if (word.contains("@") || word.contains("#") || word.contains("$")) {
return false;
}
char[] vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'};
char[] consonants = {
'b', 'c', 'd', 'f', 'g', 'h', 'j', 'k', 'l', 'm', 'n', 'p', 'q', 'r', 's', 't', 'v',
'w', 'x', 'y', 'z', 'B', 'C', 'D', 'F', 'G', 'H', 'J', 'K', 'L', 'M', 'N', 'P', 'Q',
'R', 'S', 'T', 'V', 'W', 'X', 'Y', 'Z'
};
boolean flag1 = false;
boolean flag2 = false;
for (char c : vowels) {
if (word.indexOf(c) != -1) {
flag1 = true;
break;
}
}
for (char c : consonants) {
if (word.indexOf(c) != -1) {
flag2 = true;
break;
}
}
return flag1 && flag2;
}
}