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
import java.math.BigInteger;

class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int countAnagrams(String s) {
        int n = s.length();
        long[] f = new long[n + 1];
        f[0] = 1;
        for (int i = 1; i <= n; ++i) {
            f[i] = f[i - 1] * i % MOD;
        }
        long p = 1;
        for (String w : s.split(" ")) {
            int[] cnt = new int[26];
            for (int i = 0; i < w.length(); ++i) {
                ++cnt[w.charAt(i) - 'a'];
            }
            p = p * f[w.length()] % MOD;
            for (int v : cnt) {
                p = p * BigInteger.valueOf(f[v]).modInverse(BigInteger.valueOf(MOD)).intValue()
                    % MOD;
            }
        }
        return (int) p;
    }
}