Given an array of strings strs, return the length of the longest uncommon subsequence between them. If the longest uncommon subsequence does not exist, return -1.

An uncommon subsequence between an array of strings is a string that is a subsequence of one string but not the others.

A subsequence of a string s is a string that can be obtained after deleting any number of characters from s.

• For example, "abc" is a subsequence of "aebdc" because you can delete the underlined characters in "aebdc" to get "abc". Other subsequences of "aebdc" include "aebdc", "aeb", and "" (empty string).

Example 1:

Input: strs = ["aba","cdc","eae"]
Output: 3

Example 2:

Input: strs = ["aaa","aaa","aa"]
Output: -1

Constraints:

• 2 <= strs.length <= 50
• 1 <= strs[i].length <= 10
• strs[i] consists of lowercase English letters.

impl Solution {
    pub fn find_lu_slength(strs: Vec<String>) -> i32 {
        let mut max_length = -1;

        for i in 0..strs.len() {
						// flag ~_~
            let mut is_uncommon = true;
            for j in 0..strs.len() {
                if i != j && Solution::is_subsequence(&strs[i], &strs[j]) {
                    is_uncommon = false;
                    break;
                }
            }
            if is_uncommon {
                max_length = std::cmp::max(max_length, strs[i].len() as i32);
            }
        }
        max_length
    }
    
    pub fn is_subsequence(s: &str, t: &str) -> bool {
        let mut i = 0;
        let mut j = 0;
				// 문자열을 벡터로 변환
				// why?! 러스트는 문자열을 UTF-8 로 인코딩되어 각 문자의 크기가 가변적이며,
				// 이를 직접 접근하도록 지원하는 언어도 있지만... 러스트 이놈은 직접 접근을 지원 안함
        let s_chars: Vec<char> = s.chars().collect();
        let t_chars: Vec<char> = t.chars().collect();
        while i < s_chars.len() && j < t_chars.len() {
            if s_chars[i] == t_chars[j] {
                i += 1;
            }
            j += 1;
        }
        i == s_chars.len()
    }
}