522. Longest Uncommon Subsequence II
Read the full problem statement on LeetCode.
Difficulty: medium Acceptance: 43% Topics: Array, Hash Table, Two Pointers, String, Sorting
View full problem on LeetCode Reading material
Reference solution (spoiler · python)
# Time: O(l * n^2)
# Space: O(1)
class Solution(object):
def findLUSlength(self, strs):
"""
:type strs: List[str]
:rtype: int
"""
def isSubsequence(a, b):
i = 0
for j in xrange(len(b)):
if i >= len(a):
break
if a[i] == b[j]:
i += 1
return i == len(a)
strs.sort(key=len, reverse=True)
for i in xrange(len(strs)):
all_of = True
for j in xrange(len(strs)):
if len(strs[j]) < len(strs[i]):
break
if i != j and isSubsequence(strs[i], strs[j]):
all_of = False
break
if all_of:
return len(strs[i])
return -1
Solution from kamyu104/LeetCode-Solutions · MIT
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