Problem

Given two strings a and b, return _the length of thelongest 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.

Examples

Example 1

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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

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Input: a = "aaa", b = "bbb"
Output: 3
Explanation:  The longest uncommon subsequences are "aaa" and "bbb".

Example 3

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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.

Solution

Method 1 – Simple Comparison

Intuition

If the two strings are not equal, the longer string itself is an uncommon subsequence, since it cannot be a subsequence of the other. If they are equal, there is no uncommon subsequence.

Approach

  1. Compare the two strings a and b.
  2. If they are equal, return -1 (no uncommon subsequence).
  3. Otherwise, return the length of the longer string.

Code

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class Solution {
public:
    int findLUSlength(string a, string b) {
        return a == b ? -1 : max(a.size(), b.size());
    }
};
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func findLUSlength(a string, b string) int {
    if a == b {
        return -1
    }
    if len(a) > len(b) {
        return len(a)
    }
    return len(b)
}
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class Solution {
    public int findLUSlength(String a, String b) {
        return a.equals(b) ? -1 : Math.max(a.length(), b.length());
    }
}
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class Solution {
    fun findLUSlength(a: String, b: String): Int {
        return if (a == b) -1 else maxOf(a.length, b.length)
    }
}
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class Solution:
    def findLUSlength(self, a: str, b: str) -> int:
        return -1 if a == b else max(len(a), len(b))
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impl Solution {
    pub fn find_lus_length(a: String, b: String) -> i32 {
        if a == b { -1 } else { a.len().max(b.len()) as i32 }
    }
}
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class Solution {
    findLUSlength(a: string, b: string): number {
        return a === b ? -1 : Math.max(a.length, b.length);
    }
}

Complexity

  • ⏰ Time complexity: O(n), where n is the length of the longer string, for string comparison.
  • 🧺 Space complexity: O(1), as no extra space is used.