730. Count Different Palindromic

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

Link:

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

解题方法：

参考 http://zxi.mytechroad.com/blog/dynamic-programming/leetcode-730-count-different-palindromic-subsequences/
与允许duplicate的版本相似，当左右不相等时：DP[start][end] = DP[start + 1][end] + DP[start][end - 1] - DP[start + 1][end - 1];
当左右相等时，有三种情况：
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

Tips:

每次都 + 再 %，防止数据过大

Time Complexity：

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];
}