Count Anagrams
<p>You are given a string <code>s</code> containing one or more words. Every consecutive pair of words is separated by a single space <code>&#39; &#39;</code>.</p>

<p>A string <code>t</code> is an <strong>anagram</strong> of string <code>s</code> if the <code>i<sup>th</sup></code> word of <code>t</code> is a <strong>permutation</strong> of the <code>i<sup>th</sup></code> word of <code>s</code>.</p>

<ul>
	<li>For example, <code>&quot;acb dfe&quot;</code> is an anagram of <code>&quot;abc def&quot;</code>, but <code>&quot;def cab&quot;</code>&nbsp;and <code>&quot;adc bef&quot;</code> are not.</li>
</ul>

<p>Return <em>the number of <strong>distinct anagrams</strong> of </em><code>s</code>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;too hot&quot;
<strong>Output:</strong> 18
<strong>Explanation:</strong> Some of the anagrams of the given string are &quot;too hot&quot;, &quot;oot hot&quot;, &quot;oto toh&quot;, &quot;too toh&quot;, and &quot;too oht&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aa&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> There is only one anagram possible for the given string.</pre>

<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 of lowercase English letters and spaces <code>&#39; &#39;</code>.</li>
	<li>There is single space between consecutive words.</li>
</ul>
Hash Table; Math; String; Combinatorics; Counting
class Solution {
public:
    const int mod = 1e9 + 7;

    int countAnagrams(string s) {
        stringstream ss(s);
        string w;
        long ans = 1, mul = 1;
        while (ss >> w) {
            int cnt[26] = {0};
            for (int i = 1; i <= w.size(); ++i) {
                int c = w[i - 1] - 'a';
                ++cnt[c];
                ans = ans * i % mod;
                mul = mul * cnt[c] % mod;
            }
        }
        return ans * pow(mul, mod - 2) % mod;
    }

    long pow(long x, int n) {
        long res = 1L;
        for (; n; n /= 2) {
            if (n % 2) res = res * x % mod;
            x = x * x % mod;
        }
        return res;
    }
};