# 730. Count Different Palindromic Subsequences

#### 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:

``````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.

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

https://leetcode.com/problems/count-different-palindromic-subsequences/description/

#### 解题方法：

a xxxx a
`DP[start][end] = 2 * DP[start + 1][end - 1] + 2;` 除了乘2以外，还多了a与aa
a xxxa a
`DP[start][end] = 2 * DP[start + 1][end - 1] + 1;` 多了一个aa
a axxa a
`DP[start][end] = 2 * DP[start + 1][end - 1] - DP[left + 1][right - 1];` 需要减去中间aa之间重叠的部分

O(n ^ 3)

#### 完整代码：

``````int countPalindromicSubsequences(string S) {
int len = S.size();
if(!len)
return 0;
vector<vector<int>> DP(len, vector<int>(len, 0));
long int M=1000000007;
for(int i = 0; i < len; i++)
DP[i][i] = 1;
for(int l = 2; l <= len; l++) {
for(int start = 0; start + l - 1 < len; start++) {
int end = start + l - 1;
if(S[start] == S[end]) {
int cnt = 0;
int left = start + 1;
int right = end - 1;
while(left <= right && S[left] != S[start]) left++;
while(left <= right && S[right] != S[start]) right--;
if(left > right)
DP[start][end] = 2 * DP[start + 1][end - 1] + 2;
else if(left == right)
DP[start][end] = 2 * DP[start + 1][end - 1] + 1;
else
DP[start][end] = 2 * DP[start + 1][end - 1] - DP[left + 1][right - 1];
}
else
DP[start][end] = DP[start + 1][end] + DP[start][end - 1] - DP[start + 1][end - 1];
DP[start][end] = (DP[start][end] + M) % M;
}
}
return DP[0][len - 1];
}
``````