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feat: add solutions to lc problem: No.2950 (doocs#2040)
No.2950.Number of Divisible Substrings
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‎solution/2900-2999/2950.Number of Divisible Substrings/README.md

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# [2950. Number of Divisible Substrings](https://leetcode.cn/problems/number-of-divisible-substrings)
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[English Version](/solution/2900-2999/2950.Number%20of%20Divisible%20Substrings/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>Each character of the English alphabet has been mapped to a digit as shown below.</p>
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<p><img alt="" src="https://assets.leetcode.com/uploads/2023/11/28/old_phone_digits.png" style="padding: 10px; width: 200px; height: 200px;" /></p>
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<p>A string is <strong>divisible</strong> if the sum of the mapped values of its characters is divisible by its length.</p>
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<p>Given a string <code>s</code>, return <em>the number of <strong>divisible substrings</strong> of</em> <code>s</code>.</p>
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<p>A <strong>substring</strong> is a contiguous non-empty sequence of characters within a string.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<table border="1" cellspacing="3" style="border-collapse: separate; text-align: center;">
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<tbody>
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<tr>
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<th style="padding: 5px; border: 1px solid black;">Substring</th>
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<th style="padding: 5px; border: 1px solid black;">Mapped</th>
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<th style="padding: 5px; border: 1px solid black;">Sum</th>
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<th style="padding: 5px; border: 1px solid black;">Length</th>
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<th style="padding: 5px; border: 1px solid black;">Divisible?</th>
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</tr>
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<tr>
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<td style="padding: 5px; border: 1px solid black;">a</td>
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<td style="padding: 5px; border: 1px solid black;">1</td>
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<td style="padding: 5px; border: 1px solid black;">1</td>
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<td style="padding: 5px; border: 1px solid black;">1</td>
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<td style="padding: 5px; border: 1px solid black;">Yes</td>
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</tr>
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<tr>
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<td style="padding: 5px; border: 1px solid black;">s</td>
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<td style="padding: 5px; border: 1px solid black;">7</td>
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<td style="padding: 5px; border: 1px solid black;">7</td>
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<td style="padding: 5px; border: 1px solid black;">1</td>
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<td style="padding: 5px; border: 1px solid black;">Yes</td>
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</tr>
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<tr>
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<td style="padding: 5px; border: 1px solid black;">d</td>
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<td style="padding: 5px; border: 1px solid black;">2</td>
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<td style="padding: 5px; border: 1px solid black;">2</td>
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<td style="padding: 5px; border: 1px solid black;">1</td>
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<td style="padding: 5px; border: 1px solid black;">Yes</td>
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</tr>
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<tr>
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<td style="padding: 5px; border: 1px solid black;">f</td>
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<td style="padding: 5px; border: 1px solid black;">3</td>
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<td style="padding: 5px; border: 1px solid black;">3</td>
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<td style="padding: 5px; border: 1px solid black;">1</td>
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<td style="padding: 5px; border: 1px solid black;">Yes</td>
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</tr>
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<tr>
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<td style="padding: 5px; border: 1px solid black;">as</td>
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<td style="padding: 5px; border: 1px solid black;">1, 7</td>
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<td style="padding: 5px; border: 1px solid black;">8</td>
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<td style="padding: 5px; border: 1px solid black;">2</td>
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<td style="padding: 5px; border: 1px solid black;">Yes</td>
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</tr>
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<tr>
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<td style="padding: 5px; border: 1px solid black;">sd</td>
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<td style="padding: 5px; border: 1px solid black;">7, 2</td>
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<td style="padding: 5px; border: 1px solid black;">9</td>
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<td style="padding: 5px; border: 1px solid black;">2</td>
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<td style="padding: 5px; border: 1px solid black;">No</td>
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</tr>
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<tr>
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<td style="padding: 5px; border: 1px solid black;">df</td>
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<td style="padding: 5px; border: 1px solid black;">2, 3</td>
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<td style="padding: 5px; border: 1px solid black;">5</td>
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<td style="padding: 5px; border: 1px solid black;">2</td>
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<td style="padding: 5px; border: 1px solid black;">No</td>
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</tr>
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<tr>
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<td style="padding: 5px; border: 1px solid black;">asd</td>
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<td style="padding: 5px; border: 1px solid black;">1, 7, 2</td>
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<td style="padding: 5px; border: 1px solid black;">10</td>
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<td style="padding: 5px; border: 1px solid black;">3</td>
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<td style="padding: 5px; border: 1px solid black;">No</td>
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</tr>
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<tr>
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<td style="padding: 5px; border: 1px solid black;">sdf</td>
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<td style="padding: 5px; border: 1px solid black;">7, 2, 3</td>
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<td style="padding: 5px; border: 1px solid black;">12</td>
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<td style="padding: 5px; border: 1px solid black;">3</td>
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<td style="padding: 5px; border: 1px solid black;">Yes</td>
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</tr>
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<tr>
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<td style="padding: 5px; border: 1px solid black;">asdf</td>
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<td style="padding: 5px; border: 1px solid black;">1, 7, 2, 3</td>
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<td style="padding: 5px; border: 1px solid black;">13</td>
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<td style="padding: 5px; border: 1px solid black;">4</td>
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<td style="padding: 5px; border: 1px solid black;">No</td>
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</tr>
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</tbody>
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</table>
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<pre>
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<strong>Input:</strong> word = &quot;asdf&quot;
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<strong>Output:</strong> 6
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<strong>Explanation:</strong> The table above contains the details about every substring of word, and we can see that 6 of them are divisible.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> word = &quot;bdh&quot;
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<strong>Output:</strong> 4
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<strong>Explanation:</strong> The 4 divisible substrings are: &quot;b&quot;, &quot;d&quot;, &quot;h&quot;, &quot;bdh&quot;.
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It can be shown that there are no other substrings of word that are divisible.
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> word = &quot;abcd&quot;
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<strong>Output:</strong> 6
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<strong>Explanation:</strong> The 6 divisible substrings are: &quot;a&quot;, &quot;b&quot;, &quot;c&quot;, &quot;d&quot;, &quot;ab&quot;, &quot;cd&quot;.
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It can be shown that there are no other substrings of word that are divisible.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= word.length &lt;= 2000</code></li>
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<li><code>word</code> consists only of lowercase English letters.</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：枚举**
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我们先用一个哈希表或数组 $mp$ 记录每个字母对应的数字。
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然后，我们枚举子串的起始位置 $i$，再枚举子串的结束位置 $j$，计算子串 $s[i..j]$ 的数字和 $s$，如果 $s$ 能被 $j-i+1$ 整除，那么就找到了一个可被整除的子串，将答案加一。
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枚举结束后，返回答案。
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时间复杂度 $O(n^2)$，空间复杂度 $O(C)$。其中 $n$ 是字符串 $word$ 的长度，而 $C$ 是字符集的大小，本题中 $C=26$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def countDivisibleSubstrings(self, word: str) -> int:
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d = ["ab", "cde", "fgh", "ijk", "lmn", "opq", "rst", "uvw", "xyz"]
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mp = {}
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for i, s in enumerate(d, 1):
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for c in s:
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mp[c] = i
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ans = 0
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n = len(word)
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for i in range(n):
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s = 0
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for j in range(i, n):
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s += mp[word[j]]
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ans += s % (j - i + 1) == 0
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int countDivisibleSubstrings(String word) {
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String[] d = {"ab", "cde", "fgh", "ijk", "lmn", "opq", "rst", "uvw", "xyz"};
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int[] mp = new int[26];
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for (int i = 0; i < d.length; ++i) {
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for (char c : d[i].toCharArray()) {
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mp[c - 'a'] = i + 1;
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}
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}
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int ans = 0;
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int n = word.length();
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for (int i = 0; i < n; ++i) {
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int s = 0;
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for (int j = i; j < n; ++j) {
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s += mp[word.charAt(j) - 'a'];
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ans += s % (j - i + 1) == 0 ? 1 : 0;
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int countDivisibleSubstrings(string word) {
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string d[9] = {"ab", "cde", "fgh", "ijk", "lmn", "opq", "rst", "uvw", "xyz"};
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int mp[26]{};
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for (int i = 0; i < 9; ++i) {
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for (char& c : d[i]) {
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mp[c - 'a'] = i + 1;
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}
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}
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int ans = 0;
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int n = word.size();
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for (int i = 0; i < n; ++i) {
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int s = 0;
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for (int j = i; j < n; ++j) {
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s += mp[word[j] - 'a'];
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ans += s % (j - i + 1) == 0 ? 1 : 0;
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func countDivisibleSubstrings(word string) (ans int) {
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d := []string{"ab", "cde", "fgh", "ijk", "lmn", "opq", "rst", "uvw", "xyz"}
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mp := [26]int{}
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for i, s := range d {
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for _, c := range s {
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mp[c-'a'] = i + 1
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}
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}
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n := len(word)
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for i := 0; i < n; i++ {
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s := 0
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for j := i; j < n; j++ {
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s += mp[word[j]-'a']
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if s%(j-i+1) == 0 {
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ans++
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}
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}
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}
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return
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}
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```
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### **TypeScript**
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```ts
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function countDivisibleSubstrings(word: string): number {
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const d: string[] = ['ab', 'cde', 'fgh', 'ijk', 'lmn', 'opq', 'rst', 'uvw', 'xyz'];
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const mp: number[] = Array(26).fill(0);
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for (let i = 0; i < d.length; ++i) {
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for (const c of d[i]) {
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mp[c.charCodeAt(0) - 'a'.charCodeAt(0)] = i + 1;
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}
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}
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const n = word.length;
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let ans = 0;
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for (let i = 0; i < n; ++i) {
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let s = 0;
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for (let j = i; j < n; ++j) {
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s += mp[word.charCodeAt(j) - 'a'.charCodeAt(0)];
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if (s % (j - i + 1) === 0) {
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++ans;
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}
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}
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}
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return ans;
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}
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```
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### **Rust**
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```rust
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impl Solution {
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pub fn count_divisible_substrings(word: String) -> i32 {
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let d = vec!["ab", "cde", "fgh", "ijk", "lmn", "opq", "rst", "uvw", "xyz"];
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let mut mp = vec![0; 26];
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for (i, s) in d.iter().enumerate() {
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s.chars().for_each(|c| {
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mp[(c as usize) - ('a' as usize)] = (i + 1) as i32;
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});
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}
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let mut ans = 0;
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let n = word.len();
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for i in 0..n {
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let mut s = 0;
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for j in i..n {
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s += mp[(word.as_bytes()[j] as usize) - ('a' as usize)];
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ans += (s % ((j - i + 1) as i32) == 0) as i32;
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}
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}
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ans
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}
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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