Algorithm
Problem Name: 730. Count Different Palindromic Subsequences
Given a string s, return the number of different non-empty palindromic subsequences in s. Since the answer may be very large, return it modulo 109 + 7.
A subsequence of a string is obtained by deleting zero or more characters from the string.
A sequence is palindromic if it is equal to the sequence reversed.
Two sequences a1, a2, ... and b1, b2, ... are different if there is some i for which ai != bi.
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 109 + 7.
Constraints:
1 <= s.length <= 1000s[i]is either'a','b','c', or'd'.
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
const countPalindromicSubsequences = function(S) {
const len = S.length
const dp = Array.from({ length: len }, () => new Array(len).fill(0))
const mod = 10 ** 9 + 7
for (let i = 0; i < len; i++) dp[i][i] = 1
for (let distance = 1; distance < len; distance++) {
for (let i = 0; i < len - distance; i++) {
let j = i + distance
if (S[i] === S[j]) {
let low = i + 1
let high = j - 1
while (low <= high && S[low] != S[j]) low++
while (low <= high && S[high] != S[j]) 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] = dp[i][j] < 0 ? dp[i][j] + mod : dp[i][j] % mod
}
}
return dp[0][len - 1]
}
Input
s = "bccb"
Output
6
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def countPalindromicSubsequences(self, S):
mod, memo = 10 ** 9 + 7, {}
def dfs(i, j):
if (i, j) not in memo:
cnt = 0
for x in "abcd":
try: l, r = S[i:j + 1].index(x) + i, S[i:j + 1].rindex(x) + i
except: continue
cnt += l != r and dfs(l + 1, r - 1) + 2 or 1
memo[(i, j)] = cnt % mod
return memo[(i, j)]
return dfs(0, len(S) - 1)
Input
s = "bccb"
Output
6