-
-
Notifications
You must be signed in to change notification settings - Fork 298
/
Copy path903.py
44 lines (40 loc) · 1.57 KB
/
903.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
__________________________________________________________________________________________________
sample 48 ms submission
# https://leetcode.com/problems/valid-permutations-for-di-sequence/discuss/168278/C%2B%2BJavaPython-DP-Solution-O(N2)
class Solution:
def numPermsDISequence(self, S: str) -> int:
dp = [1] * (len(S) + 1)
for c in S:
if c == "I":
dp = dp[:-1]
for i in range(1, len(dp)):
dp[i] += dp[i - 1]
else:
dp = dp[1:]
for i in range(len(dp) - 1)[::-1]:
dp[i] += dp[i + 1]
return dp[0] % (10**9 + 7)
__________________________________________________________________________________________________
sample 64 ms submission
class Solution:
def numPermsDISequence(self, S: str) -> int:
if not S:
return 0
mod = 1000000007
n = len(S)
dp = []
for i in range(n+1):
dp.append([0] * (n+1))
dp[0][0] = 1
for i in range(1, n+1):
for j in range(i+1):
if S[i-1] == 'D':
if j != 0:
dp[i][j] = (dp[i-1][i-1] - dp[i-1][j-1] + dp[i][j-1]) % mod
else:
dp[i][j] = dp[i-1][i-1]
elif S[i-1] == 'I':
if j != 0:
dp[i][j] = (dp[i-1][j-1] + dp[i][j-1]) % mod
return dp[n][n]
__________________________________________________________________________________________________