Formatted question description: https://leetcode.ca/all/730.html
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:
S will be in the range [1, 1000].S[i] will be in the set {'a', 'b', 'c', 'd'}.Related Topics:
String, Dynamic Programming
Similar Questions:
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], ifS[i] != S[j]dp[i + 1][j] + dp[i][j - 1] + 1, ifS[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 > jori == 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], ifS[i] != S[j]2 + sum( dp[i+1][j-1][t] | 0 <= t < 4 ), ifS[i] == S[j] && S[i] == 'a' + k
// OJ: https://leetcode.com/problems/count-different-palindromic-subsequences/
// 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;
}
};
Java
-
class Solution { public int countPalindromicSubsequences(String S) { final int MODULO = 1000000007; int length = S.length(); int[][] dp = new int[length][length]; for (int i = 0; i < length; i++) dp[i][i] = 1; for (int i = length - 2; i >= 0; i--) { char c1 = S.charAt(i); for (int j = i + 1; j < length; j++) { char c2 = S.charAt(j); if (c1 == c2) { int low = i + 1, high = j - 1; while (low <= high && S.charAt(low) != c1) low++; while (low <= high && S.charAt(high) != c2) high--; if (low > high) dp[i][j] = dp[i + 1][j - 1] * 2 + 2; else if (low == high) dp[i][j] = dp[i + 1][j - 1] * 2 + 1; else dp[i][j] = dp[i + 1][j - 1] * 2 - dp[low + 1][high - 1]; } else dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1]; dp[i][j] %= MODULO; if (dp[i][j] < 0) dp[i][j] += MODULO; } } return dp[0][length - 1]; } } -
// OJ: https://leetcode.com/problems/count-different-palindromic-subsequences/ // 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; } }; -
class Solution: def countPalindromicSubsequences(self, S): """ :type S: str :rtype: int """ def count(S, i, j): if i > j: return 0 if i == j: return 1 if self.m_[i][j]: return self.m_[i][j] if S[i] == S[j]: ans = count(S, i + 1, j - 1) * 2 l = i + 1 r = j - 1 while l <= r and S[l] != S[i]: l += 1 while l <= r and S[r] != S[i]: r -= 1 if l > r: ans += 2 elif l == r: ans += 1 else: ans -= count(S, l + 1, r - 1) else: ans = count(S, i + 1, j) + count(S, i, j - 1) - count(S, i + 1, j - 1) self.m_[i][j] = ans % (10 ** 9 + 7) return self.m_[i][j] n = len(S) self.m_ = [[None for _ in range(n)] for _ in range(n)] return count(S, 0, n - 1)