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2484. Count Palindromic Subsequences

Description

Given a string of digits s, return the number of palindromic subsequences of s having length 5. Since the answer may be very large, return it modulo 109 + 7.

Note:

  • A string is palindromic if it reads the same forward and backward.
  • A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

 

Example 1:

Input: s = "103301"
Output: 2
Explanation: 
There are 6 possible subsequences of length 5: "10330","10331","10301","10301","13301","03301". 
Two of them (both equal to "10301") are palindromic.

Example 2:

Input: s = "0000000"
Output: 21
Explanation: All 21 subsequences are "00000", which is palindromic.

Example 3:

Input: s = "9999900000"
Output: 2
Explanation: The only two palindromic subsequences are "99999" and "00000".

 

Constraints:

  • 1 <= s.length <= 104
  • s consists of digits.

Solutions

Solution 1: Enumeration + Counting

The time complexity is $O(100 \times n)$, and the space complexity is $O(100 \times n)$. Where $n$ is the length of the string $s$.

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class Solution:
    def countPalindromes(self, s: str) -> int:
        mod = 10**9 + 7
        n = len(s)
        pre = [[[0] * 10 for _ in range(10)] for _ in range(n + 2)]
        suf = [[[0] * 10 for _ in range(10)] for _ in range(n + 2)]
        t = list(map(int, s))
        c = [0] * 10
        for i, v in enumerate(t, 1):
            for j in range(10):
                for k in range(10):
                    pre[i][j][k] = pre[i - 1][j][k]
            for j in range(10):
                pre[i][j][v] += c[j]
            c[v] += 1
        c = [0] * 10
        for i in range(n, 0, -1):
            v = t[i - 1]
            for j in range(10):
                for k in range(10):
                    suf[i][j][k] = suf[i + 1][j][k]
            for j in range(10):
                suf[i][j][v] += c[j]
            c[v] += 1
        ans = 0
        for i in range(1, n + 1):
            for j in range(10):
                for k in range(10):
                    ans += pre[i - 1][j][k] * suf[i + 1][j][k]
                    ans %= mod
        return ans

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class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int countPalindromes(String s) {
        int n = s.length();
        int[][][] pre = new int[n + 2][10][10];
        int[][][] suf = new int[n + 2][10][10];
        int[] t = new int[n];
        for (int i = 0; i < n; ++i) {
            t[i] = s.charAt(i) - '0';
        }
        int[] c = new int[10];
        for (int i = 1; i <= n; ++i) {
            int v = t[i - 1];
            for (int j = 0; j < 10; ++j) {
                for (int k = 0; k < 10; ++k) {
                    pre[i][j][k] = pre[i - 1][j][k];
                }
            }
            for (int j = 0; j < 10; ++j) {
                pre[i][j][v] += c[j];
            }
            c[v]++;
        }
        c = new int[10];
        for (int i = n; i > 0; --i) {
            int v = t[i - 1];
            for (int j = 0; j < 10; ++j) {
                for (int k = 0; k < 10; ++k) {
                    suf[i][j][k] = suf[i + 1][j][k];
                }
            }
            for (int j =