Welcome to Subscribe On Youtube
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 <= 100aandbconsist 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 } }