# 730 Count Different Palindromic Subsequences

### Problem

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

``````class Solution {
public int countPalindromicSubsequences(String S) {
int base = 1000000007;
long[][] dp = new long[S.length()][S.length()];
for (int l = 1; l <= S.length(); l ++) {
for (int i = 0; i + l - 1 < S.length(); i ++) {
int j = i + l - 1;
if (l == 1) {
dp[i][j] = 1;
continue;
}
if (l == 2) {
dp[i][j] = 2;
continue;
}
if (S.charAt(i) == S.charAt(j)) {
int left = i + 1, right = j - 1;
while (left <= right && S.charAt(left) != S.charAt(i)) {
left ++;
}
while (left <= right && S.charAt(right) != S.charAt(i)) {
right --;
}
if (left > right) {
dp[i][j] = dp[i + 1][j - 1] * 2 + 2;
}
else if (left == right) {
dp[i][j] = dp[i + 1][j - 1] * 2 + 1;
}
else {
dp[i][j] = dp[i + 1][j - 1] * 2 - dp[left + 1][right - 1];
}
}
else {
dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1] ;
}
// dp[i][j] = dp[i][j] % base;
dp[i][j] = dp[i][j] < 0? dp[i][j] + base:dp[i][j]% base;
}
}
return (int)dp[S.length() - 1];
}
}
``````