Count Complete Substrings
<p>You are given a string <code>word</code> and an integer <code>k</code>.</p>

<p>A substring <code>s</code> of <code>word</code> is <strong>complete</strong> if:</p>

<ul>
	<li>Each character in <code>s</code> occurs <strong>exactly</strong> <code>k</code> times.</li>
	<li>The difference between two adjacent characters is <strong>at most</strong> <code>2</code>. That is, for any two adjacent characters <code>c1</code> and <code>c2</code> in <code>s</code>, the absolute difference in their positions in the alphabet is <strong>at most</strong> <code>2</code>.</li>
</ul>

<p>Return <em>the number of <strong>complete </strong>substrings of</em> <code>word</code>.</p>

<p>A <strong>substring</strong> is a <strong>non-empty</strong> contiguous sequence of characters in a string.</p>

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

<pre>
<strong>Input:</strong> word = &quot;igigee&quot;, k = 2
<strong>Output:</strong> 3
<strong>Explanation:</strong> The complete substrings where each character appears exactly twice and the difference between adjacent characters is at most 2 are: <u><strong>igig</strong></u>ee, igig<u><strong>ee</strong></u>, <u><strong>igigee</strong></u>.
</pre>

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

<pre>
<strong>Input:</strong> word = &quot;aaabbbccc&quot;, k = 3
<strong>Output:</strong> 6
<strong>Explanation:</strong> The complete substrings where each character appears exactly three times and the difference between adjacent characters is at most 2 are: <strong><u>aaa</u></strong>bbbccc, aaa<u><strong>bbb</strong></u>ccc, aaabbb<u><strong>ccc</strong></u>, <strong><u>aaabbb</u></strong>ccc, aaa<u><strong>bbbccc</strong></u>, <u><strong>aaabbbccc</strong></u>.
</pre>

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

<ul>
	<li><code>1 &lt;= word.length &lt;= 10<sup>5</sup></code></li>
	<li><code>word</code> consists only of lowercase English letters.</li>
	<li><code>1 &lt;= k &lt;= word.length</code></li>
</ul>
Hash Table; String; Sliding Window
class Solution {
    public int countCompleteSubstrings(String word, int k) {
        int n = word.length();
        int ans = 0;
        for (int i = 0; i < n;) {
            int j = i + 1;
            while (j < n && Math.abs(word.charAt(j) - word.charAt(j - 1)) <= 2) {
                ++j;
            }
            ans += f(word.substring(i, j), k);
            i = j;
        }
        return ans;
    }

    private int f(String s, int k) {
        int m = s.length();
        int ans = 0;
        for (int i = 1; i <= 26 && i * k <= m; ++i) {
            int l = i * k;
            int[] cnt = new int[26];
            for (int j = 0; j < l; ++j) {
                ++cnt[s.charAt(j) - 'a'];
            }
            Map<Integer, Integer> freq = new HashMap<>();
            for (int x : cnt) {
                if (x > 0) {
                    freq.merge(x, 1, Integer::sum);
                }
            }
            if (freq.getOrDefault(k, 0) == i) {
                ++ans;
            }
            for (int j = l; j < m; ++j) {
                int a = s.charAt(j) - 'a';
                int b = s.charAt(j - l) - 'a';
                freq.merge(cnt[a], -1, Integer::sum);
                ++cnt[a];
                freq.merge(cnt[a], 1, Integer::sum);

                freq.merge(cnt[b], -1, Integer::sum);
                --cnt[b];
                freq.merge(cnt[b], 1, Integer::sum);
                if (freq.getOrDefault(k, 0) == i) {
                    ++ans;
                }
            }
        }
        return ans;
    }
}