# Count Different Palindromic Subsequences Problem

## Description

LeetCode Problem 730.

Given a string s, return the number of different non-empty palindromic subsequences in s. Since the answer may be very large, return it modulo 10^9 + 7.

A subsequence of a string is obtained by deleting zero or more characters from the string. 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:

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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:

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

Constraints:

• 1 <= s.length <= 1000
• s[i] is either ‘a’, ‘b’, ‘c’, or ‘d’.

## Sample C++ Code

Let dp[len][i][x] be the number of distinct palindromes of the subtring starting at i of length len, where the first (and last) character is x. The DP formula is:

• If s[i] != x, then dp[len][i][x] = dp[len-1][i+1][x] (ignoring the first character in this window)
• If s[i+len-1] != x then dp[len][i][x] = dp[len-1][i][x] (ignoring the last character in this window)
• If both the first and last characters are x, then we need to count the number of distinct palindromes in the sub-window from i+1 to i+len-2. Need to be careful with how we count empty string.

Since we only need the subproblems of length len-1, len-2, we only need to remember the solutions for the subproblems of length len, len-1, len-2. This is needed to pass the max test case.

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class Solution {
public:
int countPalindromicSubsequences(string s) {
int md = 1000000007;
int n = s.size();
int dp[n];
for (int len = 1; len <=n; ++len) {
for (int i = 0; i + len <=n; ++i)
for (int x = 0; x < 4; ++x)  {
int &ans = dp[i][x];
ans = 0;
int j = i + len - 1;
char c = 'a' + x;
if (len == 1)
ans = s[i] == c;
else {
if (s[i] != c)
ans = dp[i+1][x];
else if (s[j] != c)
ans = dp[i][x];
else {
ans = 2;
if (len > 2)
for (int y = 0; y < 4;++y) {
ans += dp[i+1][y];
ans %=md;
}
}
}
}
for (int i=0;i<2;++i)
for (int j = 0; j < n; ++j)
for (int x=0; x < 4;++x)
dp[i][j][x] = dp[i+1][j][x];
}
int ret = 0;
for (int x = 0; x < 4;++x)
ret = (ret + dp[x]) % md;
return ret;
}
};
``````