## Description

Given a string S, find the number of different non-empty palindromic subsequences in S, and return that number modulo 10^9 + 7.

A subsequence of a string S is obtained by deleting 0 or more characters from S.

A sequence is palindromic if it is equal to the sequence reversed.

Two sequences A_1, A_2, ... and B_1, B_2, ... are different if there is some i for which A_i != B_i.

Example 1:

Input:
S = 'bccb'
Output: 6
Explanation:
The 6 different non-empty palindromic subsequences are 'b', 'c', 'bb', 'cc', 'bcb', 'bccb'.
Note that 'bcb' is counted only once, even though it occurs twice.


Example 2:

Input:
Output: 104860361
Explanation:
There are 3104860382 different non-empty palindromic subsequences, which is 104860361 modulo 10^9 + 7.


Note:

The length of S will be in the range [1, 1000].

Each character S[i] will be in the set {'a', 'b', 'c', 'd'}.

## Solutions

### 1. 记忆化递归

# Time complexity: O(n^2)
# Space complexity: O(n^2)
class Solution:
def countPalindromicSubsequences(self, S: str) -> int:
n = len(S)
self.m_ = [[None for _ in range(n)] for _ in range(n)]
return self.count(S, 0, n - 1)

def count(self, S, i, j):
if i > j: return 0
if i == j: return 1
if self.m_[i][j]: return self.m_[i][j]

if S[i] == S[j]:
ans = self.count(S, i + 1, j - 1) * 2
l = i + 1
r = j - 1
while l <= r and S[l] != S[i]: l += 1
while l <= r and S[r] != S[i]: r -= 1
if l > r: ans += 2
elif l == r: ans += 1
else: ans -= self.count(S, l + 1, r - 1)
else:
ans = self.count(S, i + 1, j) + self.count(S, i, j - 1) - self.count(S, i + 1, j - 1)

self.m_[i][j] = ans % 1000000007
return self.m_[i][j]
# 366/366 cases passed (2208 ms)
# Your runtime beats 39.86 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (31.3 MB)


### 2. DP

class Solution:
def countPalindromicSubsequences(self, S):
n = len(S)
if n == 0: return 0
dp = [[0 for _ in range(n)] for _ in range(n)]
for i in range(n): dp[i][i] = 1

for size in range(2, n + 1):
for i in range(n - size + 1):
j = i + size - 1
if S[i] == S[j]:
dp[i][j] = dp[i + 1][j - 1] * 2
l = i + 1
r = j - 1
while l <= r and S[l] != S[i]: l += 1
while l <= r and S[r] != S[i]: r -= 1
if l > r: dp[i][j] += 2
elif l == r: dp[i][j] += 1
else: dp[i][j] -= dp[l + 1][r - 1]
else:
dp[i][j] = dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1]
dp[i][j] %= 1000000007

return dp[0][n - 1]
# 366/366 cases passed (2124 ms)
# Your runtime beats 40.56 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (36.7 MB)