730

730. Count Different Palindromic Subsequences (Hard)

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: 
S = 'abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba'
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'}.

Related Topics:
String, Dynamic Programming

Similar Questions:

• Longest Palindromic Subsequence (Medium)

Solution 1. DP

First consider the case where we count duplicates as well.

Let dp[i][j] be the number of palindromic subsequences in S[i..j].

dp[i][j] = 0   if i > j
dp[i][i] = 1

If S[i] != S[j]:

dp[i][j] = dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1]

We need to - dp[i + 1][j - 1] because the palindromic subsequences are counted twice already in dp[i + 1][j] and dp[i][j - 1].

If S[i] == S[j], then there are additional dp[i + 1][j - 1] + 1 cases where are the palindromic subsequences in S[(i+1)..(j-1)] wrapped with S[i] and S[j], plus one case S[i]S[j].

dp[i][j] = dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1] + (dp[i + 1][j - 1] + 1)
         = dp[i + 1][j] + dp[i][j - 1] + 1

So in sum, dp[i][j] = :

• 0, if i > j
• 1, if i == j
• dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1], if S[i] != S[j]
• dp[i + 1][j] + dp[i][j - 1] + 1, if S[i] == S[j]

Now consider distinct count.

dp[i][j][k] is the number of distinct palindromic subsequences in S[i..j] bordered by 'a' + k.

If S[i] != S[j]:

dp[i][j][k] = dp[i+1][j][k] + dp[i][j-1][k] - dp[i+1][j-1][k]

If S[i] == S[j] && S[i] == 'a' + k:

dp[i][j][k] = 2 + sum( dp[i+1][j-1][t] | 0 <= t < 4 )

This is because we can wrap all the cases of dp[i+1][j-1][t] with S[i] and S[j] to form new palindromes (which won't contain a and aa), and the +2 means a and aa.

So in sum, dp[i][j][k] =:

• 0, if i > j or i == j && S[i] != 'a' + k
• 1, if i == j && S[i] == 'a' + k
• dp[i+1][j][k] + dp[i][j-1][k] - dp[i+1][j-1][k], if S[i] != S[j]
• 2 + sum( dp[i+1][j-1][t] | 0 <= t < 4 ), if S[i] == S[j] && S[i] == 'a' + k

// OJ: https://leetcode.com/problems/count-different-palindromic-subsequences/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N^2)
// Ref: https://leetcode.com/problems/count-different-palindromic-subsequences/discuss/272297/DP-C%2B%2B-Clear-solution-explained
int memo[1001][1001][4];
class Solution {
    int mod = 1e9 + 7;
    string s;
    int dp(int first, int last, int ch) {
        if (first > last) return 0;
        if (first == last) return s[first] - 'a' == ch;
        if (memo[first][last][ch] != -1) return memo[first][last][ch];
        int ans = 0;
        if (s[first] == s[last] && s[first] - 'a' == ch) {
            ans = 2;
            for (int i = 0; i < 4; ++i) ans = (ans + dp(first + 1, last - 1, i)) % mod;
        } else {
            ans = (ans + dp(first, last - 1, ch)) % mod;
            ans = (ans + dp(first + 1, last, ch)) % mod;
            ans = (ans - dp(first + 1, last - 1, ch)) % mod;
            if (ans < 0) ans += mod;
        }
        return memo[first][last][ch] = ans;
    }
public:
    int countPalindromicSubsequences(string S) {
        s = S;
        memset(memo, -1, sizeof(memo));
        int ans = 0;
        for (int i = 0; i < 4; ++i) ans = (ans + dp(0, S.size() - 1, i)) % mod;
        return ans;
    }
};