Count Beautiful Substrings II

You are given a string s and a positive integer k.

Let vowels and consonants be the number of vowels and consonants in a string.

A string is beautiful if:

vowels == consonants.
(vowels * consonants) % k == 0, in other terms the multiplication of vowels and consonants is divisible by k.

Return the number of non-empty beautiful substrings in the given string s.

A substring is a contiguous sequence of characters in a string.

Vowel letters in English are 'a', 'e', 'i', 'o', and 'u'.

Consonant letters in English are every letter except vowels.

Example 1:

Input: s = "baeyh", k = 2
Output: 2
Explanation: There are 2 beautiful substrings in the given string.
- Substring "baeyh", vowels = 2 (["a",e"]), consonants = 2 (["y","h"]).
You can see that string "aeyh" is beautiful as vowels == consonants and vowels * consonants % k == 0.
- Substring "baeyh", vowels = 2 (["a",e"]), consonants = 2 (["b","y"]).
You can see that string "baey" is beautiful as vowels == consonants and vowels * consonants % k == 0.
It can be shown that there are only 2 beautiful substrings in the given string.

Example 2:

Input: s = "abba", k = 1
Output: 3
Explanation: There are 3 beautiful substrings in the given string.
- Substring "abba", vowels = 1 (["a"]), consonants = 1 (["b"]).
- Substring "abba", vowels = 1 (["a"]), consonants = 1 (["b"]).
- Substring "abba", vowels = 2 (["a","a"]), consonants = 2 (["b","b"]).
It can be shown that there are only 3 beautiful substrings in the given string.

Example 3:

Input: s = "bcdf", k = 1
Output: 0
Explanation: There are no beautiful substrings in the given string.

Constraints:

1 <= s.length <= 5 * 10⁴
1 <= k <= 1000
s consists of only English lowercase letters.

解题思路

这题难的地方是在数论的部分，把 $x^2 \equiv 0 \pmod{k}$，找到一个 $m$ 等价变成 $x \equiv 0 \pmod{m}$ 还是很难想到的。

还是很容易知道是哈希表维护统计那一类题目。先将字符串构造出一个序列 $a$，如果 $\text{str}_i$ 是元音字母，则 $a_i = 1$，否则 $\text{str}_i$ 是辅音字母，则 $a_i = -1$。然后对序列 $a$ 求前缀和得到 $s$，当枚举到右端点 $r$，那么子数组的左端点 $l \in [0, r - 1]$ 就要满足 $s_r - s_l = 0$，即在左边找到满足 $s_l = s_r$ 的 $l$ 的数量。

还要满足另外一个条件，$\left(\frac{r-l}{2}\right)^2 \equiv 0 \pmod{k}$，式子等价于 $\left(\frac{r-l}{2}\right)^2 = x \cdot k, \, x \in \mathbb{Z}$。继续推有

\begin{align*}
\left(\frac{r-l}{2}\right)^2 &= x \cdot k \\
(r-l)^2 &= 4 x \cdot k \\
(r - l)^2 &\equiv 0 \pmod{4k}
\end{align*}

然后想到说能不能把左式的平方去掉，即找到一个 $m$，使得 $r - l \equiv 0 \pmod{m}$ 与上式的同余方程是等价的。对 $4k$ 进行质因数分解得到 $4k = p_1^{\alpha{1}} \cdot p_2^{\alpha{2}} \cdots p_n^{\alpha{n}}$，那么很明显对于每一项 $p_i^{\alpha{i}}$，$r-l$ 的质因数分解也必须含有 $p_i^{\beta{i}}$ 且 $2 \cdot \beta{i} \geq \alpha{i}$，即 $\beta{i} \geq \left\lceil \frac{\alpha{i}}{2} \right\rceil$。因此有 $m = p_1^{\left\lceil \frac{\alpha{1}}{2} \right\rceil} \cdot p_2^{\left\lceil \frac{\alpha{2}}{2} \right\rceil} \cdots p_n^{\left\lceil \frac{\alpha{n}}{2} \right\rceil}$，那么两个同余方程就可以等价了。

因此把哈希表开成两维，当枚举完右端点 $r$，维护 $\text{mp}[s_r][r \bmod m] \gets \text{mp}[s_r][r \bmod m] + 1$ 即可。由于 $s_r$ 可能是负数所以需要加上一个偏移量 $n$。

AC 代码如下，时间复杂度为 $O(\sqrt{k} + n)$：

class Solution {
public:
    long long beautifulSubstrings(string str, int k) {
        int n = str.size();
        vector<int> s(n + 1);
        set<char> st({'a', 'e', 'i', 'o', 'u'});
        for (int i = 1; i <= n; i++) {
            s[i] = s[i - 1];
            if (st.count(str[i - 1])) s[i]++;
            else s[i]--;
        }
        int m = 1;
        k <<= 2;
        for (int i = 2; i * i <= k; i++) {
            if (k % i == 0) {
                int s = 0;
                while (k % i == 0) {
                    s++;
                    k /= i;
                }
                s = s + 1 >> 1;
                while (s--) {
                    m *= i;
                }
            }
        }
        if (k > 1) m *= k;
        vector<vector<int>> mp(n * 2 + 1, vector<int>(m));
        long long ret = 0;
        for (int i = 1; i <= n; i++) {
            mp[s[i - 1] + n][(i - 1) % m]++;
            ret += mp[s[i] + n][i % m];
        }
        
        return ret;
    }
};