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Tested Python solution for LeetCode 521 with 22 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 521 # by problem number
lcpy gen -s longest_uncommon_subsequence_i # by problem name
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.
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.
Input: a = "aaa", b = "bbb"
Output: 3
Explanation: The longest uncommon subsequences are "aaa" and "bbb".
Input: a = "aaa", b = "aaa"
Output: -1
Explanation: Every subsequence of string `a` is also a subsequence of string `b`.
a and b consist of lower-case English letters.class Solution:
# Time: O(min(len(a), len(b)))
# Space: O(1)
def find_luslength(self, a: str, b: str) -> int:
# A whole string is always a subsequence of itself, so if a != b the
# longer (or either, on a tie) string cannot be a subsequence of the
# other: equal-length subsequences imply equality. If a == b every
# subsequence is shared, so nothing is uncommon.
return max(len(a), len(b)) if a != b else -1
| Time | Space |
|---|---|
| O(min(len(a), len(b))) | O(1) |