LeetCode 0521. Longest Uncommon Subsequence I Solution in Java, C++, Python & More | Explanation + Code

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).

PythonJavaC++GoTypeScriptRust

class Solution: def findLUSlength(self, a: str, b: str) -> int: return -1 if a == b else max(len(a), len(b))(code-box)

class Solution { public int findLUSlength(String a, String b) { return a.equals(b) ? -1 : Math.max(a.length(), b.length()); } }(code-box)

class Solution { public: int findLUSlength(string a, string b) { return a == b ? -1 : max(a.size(), b.size()); } };(code-box)

func findLUSlength(a string, b string) int { if a == b { return -1 } if len(a) > len(b) { return len(a) } return len(b) }(code-box)

function findLUSlength(a: string, b: string): number { return a === b ? -1 : Math.max(a.length, b.length); }(code-box)

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