forked from kamyu104/LeetCode-Solutions
-
Notifications
You must be signed in to change notification settings - Fork 0
/
find-longest-special-substring-that-occurs-thrice-i.py
47 lines (44 loc) · 1.33 KB
/
find-longest-special-substring-that-occurs-thrice-i.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
# Time: O(26 * 3 + n * 3)
# Space: O(26 * 3)
# string, hash table
class Solution(object):
def maximumLength(self, s):
"""
:type s: str
:rtype: int
"""
lookup = [[0]*3 for _ in xrange(26)]
result = cnt = 0
for i, c in enumerate(s):
cnt += 1
if i+1 != len(s) and s[i+1] == s[i]:
continue
curr = lookup[ord(c)-ord('a')]
for j in xrange(len(curr)):
if curr[j] < cnt:
cnt, curr[j] = curr[j], cnt
cnt = 0
result = max(result, max(curr[0]-2, min(curr[0]-1, curr[1]), curr[2]))
return result if result else -1
# Time: O(26 + n^2)
# Space: O(26 + n^2)
# string, brute force, freq table
class Solution2(object):
def maximumLength(self, s):
"""
:type s: str
:rtype: int
"""
lookup = [[0] for _ in xrange(26)]
result = 0
for i, c in enumerate(s):
curr = lookup[ord(c)-ord('a')]
for j in xrange(i, len(s)):
if s[j] != s[i]:
break
if j-i+1 == len(curr):
curr.append(0)
curr[j-i+1] += 1
if curr[j-i+1] == 3:
result = max(result, j-i+1)
return result if result else -1