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2484. Count Palindromic Subsequences

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.

  

   

class Solution {
    public int countPalindromes(String s) {
        // 计算左侧的两位排列个数
        int[][][] l2r = countLeftToRight(s);
        // 计算右侧的两位排列个数
        int[][][] r2l = countRightToLeft(s);
        int MOD = 1000_000_007;
        long result = 0;
        // 从第3个元素开始，计算左右侧100中情况的count，相加
        for(int i = 2; i < s.length() - 2; i++) {
            for(int j = 0; j < 10; j++) {
                for(int k = 0; k < 10; k++) {
                    result += 1l * l2r[i - 1][j][k] * r2l[i + 1][j][k];
                    result = result % MOD;
                }
            }
        }
        return (int)result;
    }

    private int[][][] countLeftToRight (String s) {
        // 记录当前某数字出现的次数
        int[] cnt = new int[10];
        int[][][] count = new int[s.length()][10][10];
        // i为当前第二个数字的位置
        for(int i = 0; i < s.length(); i++) {
            int c = s.charAt(i) - '0';
            if(i > 0) {
                // j 为第一个数字 0～9
                for(int j = 0; j < 10; j++) {
                    // k 为第二个数字 0～9
                    for(int k = 0; k < 10; k++) {
                        count[i][j][k] = count[i - 1][j][k];
                        // 如果k是当前数字，那么当前的count要加上
                        if(c == k) count[i][j][k] += cnt[j];
                    }
                }
            }
            cnt[c]++;
        }
        return count;
    }

    private int[][][] countRightToLeft (String s) {
        // 记录当前某数字出现的次数
        int[] cnt = new int[10];
        int[][][] count = new int[s.length()][10][10];
        for(int i = s.length() - 1; i >= 0; i--) {
            int c = s.charAt(i) - '0';
            if(i < s.length() - 1) {
                for(int j = 0; j < 10; j++) {
                    for(int k = 0; k < 10; k++) {
                        count[i][j][k] = count[i + 1][j][k];
                        if(c == k) count[i][j][k] += cnt[j];
                    }
                }
            }
            cnt[c]++;
        }
        return count;
    }
}