Problem
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.
Solution (Java)
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());
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).