Easy
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
.
"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.public class Solution {
/*
* The gotcha point of this question is:
* 1. if a and b are identical, then there will be no common subsequence, return -1
* 2. else if a and b are of equal length, then any one of them will be a subsequence of the other string
* 3. else if a and b are of different length, then the longer one is a required subsequence because
* the longer string cannot be a subsequence of the shorter one
* Or in other words, when a.length() != b.length(), no subsequence of b will be equal to a,
* so return Math.max(a.length(), b.length())
*/
public int findLUSlength(String a, String b) {
if (a.equals(b)) {
return -1;
}
return Math.max(a.length(), b.length());
}
}