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longest-uncommon-subsequence-i.py
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longest-uncommon-subsequence-i.py
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# Time: O(min(a, b))
# Space: O(1)
# Given a group of two strings, you need to find the longest uncommon subsequence of this group of two strings.
# The longest uncommon subsequence is defined as the longest subsequence of one of these strings
# and this subsequence should not be any subsequence of the other strings.
#
# A subsequence is a sequence that can be derived from one sequence
# by deleting some characters without changing the order of the remaining elements.
# Trivially, any string is a subsequence of itself and an empty string is a subsequence of any string.
#
# The input will be two strings, and the output needs to be the length of the longest uncommon subsequence.
# If the longest uncommon subsequence doesn't exist, return -1.
#
# Example 1:
# Input: "aba", "cdc"
# Output: 3
# Explanation: The longest uncommon subsequence is "aba" (or "cdc"),
# because "aba" is a subsequence of "aba",
# but not a subsequence of any other strings in the group of two strings.
# Note:
#
# Both strings' lengths will not exceed 100.
# Only letters from a ~ z will appear in input strings.
# V0
# V1
# https://blog.csdn.net/fuxuemingzhu/article/details/79398536
class Solution(object):
def findLUSlength(self, a, b):
"""
:type a: str
:type b: str
:rtype: int
"""
return max(len(a), len(b)) if a != b else -1
# V2
# Time: O(min(a, b))
# Space: O(1)
class Solution(object):
def findLUSlength(self, a, b):
"""
:type a: str
:type b: str
:rtype: int
"""
if a == b:
return -1
return max(len(a), len(b))