Total Characters in String After Transformations I
<p>You are given a string <code>s</code> and an integer <code>t</code>, representing the number of <strong>transformations</strong> to perform. In one <strong>transformation</strong>, every character in <code>s</code> is replaced according to the following rules:</p>

<ul>
	<li>If the character is <code>&#39;z&#39;</code>, replace it with the string <code>&quot;ab&quot;</code>.</li>
	<li>Otherwise, replace it with the <strong>next</strong> character in the alphabet. For example, <code>&#39;a&#39;</code> is replaced with <code>&#39;b&#39;</code>, <code>&#39;b&#39;</code> is replaced with <code>&#39;c&#39;</code>, and so on.</li>
</ul>

<p>Return the <strong>length</strong> of the resulting string after <strong>exactly</strong> <code>t</code> transformations.</p>

<p>Since the answer may be very large, return it <strong>modulo</strong><!-- notionvc: eb142f2b-b818-4064-8be5-e5a36b07557a --> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;abcyy&quot;, t = 2</span></p>

<p><strong>Output:</strong> <span class="example-io">7</span></p>

<p><strong>Explanation:</strong></p>

<ul>
	<li><strong>First Transformation (t = 1)</strong>:

    <ul>
    	<li><code>&#39;a&#39;</code> becomes <code>&#39;b&#39;</code></li>
    	<li><code>&#39;b&#39;</code> becomes <code>&#39;c&#39;</code></li>
    	<li><code>&#39;c&#39;</code> becomes <code>&#39;d&#39;</code></li>
    	<li><code>&#39;y&#39;</code> becomes <code>&#39;z&#39;</code></li>
    	<li><code>&#39;y&#39;</code> becomes <code>&#39;z&#39;</code></li>
    	<li>String after the first transformation: <code>&quot;bcdzz&quot;</code></li>
    </ul>
    </li>
    <li><strong>Second Transformation (t = 2)</strong>:
    <ul>
    	<li><code>&#39;b&#39;</code> becomes <code>&#39;c&#39;</code></li>
    	<li><code>&#39;c&#39;</code> becomes <code>&#39;d&#39;</code></li>
    	<li><code>&#39;d&#39;</code> becomes <code>&#39;e&#39;</code></li>
    	<li><code>&#39;z&#39;</code> becomes <code>&quot;ab&quot;</code></li>
    	<li><code>&#39;z&#39;</code> becomes <code>&quot;ab&quot;</code></li>
    	<li>String after the second transformation: <code>&quot;cdeabab&quot;</code></li>
    </ul>
    </li>
    <li><strong>Final Length of the string</strong>: The string is <code>&quot;cdeabab&quot;</code>, which has 7 characters.</li>

</ul>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;azbk&quot;, t = 1</span></p>

<p><strong>Output:</strong> <span class="example-io">5</span></p>

<p><strong>Explanation:</strong></p>

<ul>
	<li><strong>First Transformation (t = 1)</strong>:

    <ul>
    	<li><code>&#39;a&#39;</code> becomes <code>&#39;b&#39;</code></li>
    	<li><code>&#39;z&#39;</code> becomes <code>&quot;ab&quot;</code></li>
    	<li><code>&#39;b&#39;</code> becomes <code>&#39;c&#39;</code></li>
    	<li><code>&#39;k&#39;</code> becomes <code>&#39;l&#39;</code></li>
    	<li>String after the first transformation: <code>&quot;babcl&quot;</code></li>
    </ul>
    </li>
    <li><strong>Final Length of the string</strong>: The string is <code>&quot;babcl&quot;</code>, which has 5 characters.</li>

</ul>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists only of lowercase English letters.</li>
	<li><code>1 &lt;= t &lt;= 10<sup>5</sup></code></li>
</ul>
Hash Table; Math; String; Dynamic Programming; Counting
class Solution {
public:
    int lengthAfterTransformations(string s, int t) {
        const int mod = 1e9 + 7;
        vector<vector<int>> f(t + 1, vector<int>(26, 0));

        for (char c : s) {
            f[0][c - 'a']++;
        }

        for (int i = 1; i <= t; ++i) {
            f[i][0] = f[i - 1][25] % mod;
            f[i][1] = (f[i - 1][0] + f[i - 1][25]) % mod;
            for (int j = 2; j < 26; ++j) {
                f[i][j] = f[i - 1][j - 1] % mod;
            }
        }

        int ans = 0;
        for (int j = 0; j < 26; ++j) {
            ans = (ans + f[t][j]) % mod;
        }

        return ans;
    }
};