comments | difficulty | edit_url | tags | |
---|---|---|---|---|
true |
Easy |
|
Given two strings a
and b
, return the length of the longest uncommon subsequence between a
and b
. If no such uncommon subsequence exists, return -1
.
An uncommon subsequence between two strings is a string that is a subsequence of exactly one of them.
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. So the answer would be -1
.
Constraints:
1 <= a.length, b.length <= 100
a
andb
consist of lower-case English letters.
class Solution:
def findLUSlength(self, a: str, b: str) -> int:
return -1 if a == b else max(len(a), len(b))
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());
}
};
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
}
}