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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;
}
};
-
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]; } } ############ class Solution { private final int MOD = (int) 1e9 + 7; public int countPalindromicSubsequences(String s) { int n = s.length(); long[][][] dp = new long[n][n][4]; for (int i = 0; i < n; ++i) { dp[i][i][s.charAt(i) - 'a'] = 1; } for (int l = 2; l <= n; ++l) { for (int i = 0; i + l <= n; ++i) { int j = i + l - 1; for (char c = 'a'; c <= 'd'; ++c) { int k = c - 'a'; if (s.charAt(i) == c && s.charAt(j) == c) { dp[i][j][k] = 2 + dp[i + 1][j - 1][0] + dp[i + 1][j - 1][1] + dp[i + 1][j - 1][2] + dp[i + 1][j - 1][3]; dp[i][j][k] %= MOD; } else if (s.charAt(i) == c) { dp[i][j][k] = dp[i][j - 1][k]; } else if (s.charAt(j) == c) { dp[i][j][k] = dp[i + 1][j][k]; } else { dp[i][j][k] = dp[i + 1][j - 1][k]; } } } } long ans = 0; for (int k = 0; k < 4; ++k) { ans += dp[0][n - 1][k]; } return (int) (ans % MOD); } } -
// 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: str) -> int: mod = 10**9 + 7 n = len(s) dp = [[[0] * 4 for _ in range(n)] for _ in range(n)] for i, c in enumerate(s): dp[i][i][ord(c) - ord('a')] = 1 for l in range(2, n + 1): for i in range(n - l + 1): j = i + l - 1 for c in 'abcd': k = ord(c) - ord('a') if s[i] == s[j] == c: dp[i][j][k] = 2 + sum(dp[i + 1][j - 1]) elif s[i] == c: dp[i][j][k] = dp[i][j - 1][k] elif s[j] == c: dp[i][j][k] = dp[i + 1][j][k] else: dp[i][j][k] = dp[i + 1][j - 1][k] return sum(dp[0][-1]) % mod ############ 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) -
func countPalindromicSubsequences(s string) int { mod := int(1e9) + 7 n := len(s) dp := make([][][]int, n) for i := range dp { dp[i] = make([][]int, n) for j := range dp[i] { dp[i][j] = make([]int, 4) } } for i, c := range s { dp[i][i][c-'a'] = 1 } for l := 2; l <= n; l++ { for i := 0; i+l <= n; i++ { j := i + l - 1 for _, c := range [4]byte{'a', 'b', 'c', 'd'} { k := int(c - 'a') if s[i] == c && s[j] == c { dp[i][j][k] = 2 + (dp[i+1][j-1][0]+dp[i+1][j-1][1]+dp[i+1][j-1][2]+dp[i+1][j-1][3])%mod } else if s[i] == c { dp[i][j][k] = dp[i][j-1][k] } else if s[j] == c { dp[i][j][k] = dp[i+1][j][k] } else { dp[i][j][k] = dp[i+1][j-1][k] } } } } ans := 0 for _, v := range dp[0][n-1] { ans += v } return ans % mod }