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 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 and b consist of lower-case English letters.

Solutions

Solution 1: Quick Thinking

If strings a and b are equal, then they have no special sequences, return -1; otherwise, return the length of the longer string.

The time complexity is $O(n)$, where $n$ is the length of the longer string among a and b. The space complexity is $O(1)$.

Python3JavaC++GoTypeScriptRust

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 ? -1 : Math.max(a.length, b.length);
}

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

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