lc 730 Count Different Palindromic Subsequences
730 Count Different Palindromic Subsequences
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‘}.
带记忆数组 Accepted
虽然题目只要求四个字母，但我们扩展普遍性，这里就做二十六个字母的。带记忆数组和动态规划的本质是差不多的。带记忆数组memo的递归解法，这种解法的思路是一层一层剥洋葱，比如"bccb"，按照字母来剥，先剥字母b，确定最外层"b _ _ b"，这会产生两个回文子序列"b"和"bb"，然后递归进中间的部分，把中间的回文子序列个数算出来加到结果res中，然后开始剥字母c，找到最外层"cc"，此时会产生两个回文子序列"c"和"cc"，然后由于中间没有字符串了，所以递归返回0，按照这种方法就可以算出所有的回文子序列了。
class Solution {
public:
int countPalindromicSubsequences(string S) {
int len = S.size();
vector<vector<int>> dp(len+1, vector<int>(len+1, 0));
vector<vector<int>> ch(26, vector<int>());
for (int i = 0; i < len; i++) {
ch[S[i]-‘a‘].push_back(i);
}
return calc(S, ch, dp, 0, len);
}
int calc(string S, vector<vector<int>>& ch, vector<vector<int>>& dp, int start, int end) {
if (start >= end) return 0;
if (dp[start][end] > 0) return dp[start][end];
long ans = 0;
for (int i = 0; i < 26; i++) {
if (ch[i].empty()) continue;
auto new_start = lower_bound(ch[i].begin(), ch[i].end(), start);
auto new_end = lower_bound(ch[i].begin(), ch[i].end(), end) - 1;
if (new_start == ch[i].end() || *new_start >= end) continue;
ans++;
if (new_start != new_end) ans++;
ans += calc(S, ch, dp, *new_start+1, *new_end);
}
dp[start][end] = ans % int(1e9+7);
return dp[start][end];
}
};