# 521. Longest Uncommon Subsequence I

## Description

Given two strings a and b, return the length of the longest uncommon subsequence between a and b. If the longest uncommon subsequence does not exist, return -1.

An uncommon subsequence between two strings is a string that is a subsequence of one but not the other.

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: a = "aba", b = "cdc"
Output: 3
Explanation: One longest uncommon subsequence is "aba" because "aba" is a subsequence of "aba" but not "cdc".
Note that "cdc" is also a longest uncommon subsequence.

Example 2:

Input: a = "aaa", b = "bbb"
Output: 3
Explanation: The longest uncommon subsequences are "aaa" and "bbb".

Example 3:

Input: a = "aaa", b = "aaa"
Output: -1
Explanation: Every subsequence of string a is also a subsequence of string b. Similarly, every subsequence of string b is also a subsequence of string a.

Constraints:

• 1 <= a.length, b.length <= 100
• a and b consist of lower-case English letters.

## Solutions

• class Solution {
public int findLUSlength(String a, String b) {
return a.equals(b) ? -1 : Math.max(a.length(), b.length());
}
}

• class Solution {
public:
int findLUSlength(string a, string b) {
return a == b ? -1 : max(a.size(), b.size());
}
};

• class Solution:
def findLUSlength(self, a: str, b: str) -> int:
return -1 if a == b else max(len(a), len(b))

• func findLUSlength(a string, b string) int {
if a == b {
return -1
}
if len(a) > len(b) {
return len(a)
}
return len(b)
}

• function findLUSlength(a: string, b: string): number {
return a != b ? Math.max(a.length, b.length) : -1;
}

• impl Solution {
pub fn find_lu_slength(a: String, b: String) -> i32 {
if a == b {
return -1;
}
a.len().max(b.len()) as i32
}
}