<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>

class Solution:
    def countCompleteSubstrings(self, word: str, k: int) -> int:
        groups = [""]
        for c in word:
            if not groups[-1]:
                groups[-1] += c
                continue
            if abs(ord(c) - ord(groups[-1][-1])) > 2:
                groups.append(c)
                continue
            groups[-1] += c
        
        def solve_size(group, size):
            total = 0
            filled = set()
            sliding = Counter(group[:size])
            for l in sliding:
                if sliding[l] == k: filled.add(l)
            for idx in range(size, len(group)):
                if len(filled) == (size // k): 
                    total += 1
                sliding[group[idx]] += 1
                sliding[group[idx - size]] -= 1

                if sliding[group[idx]] == k:
                    filled.add(group[idx])
                else: filled.discard(group[idx])
                if sliding[group[idx - size]] == k:
                    filled.add(group[idx - size])
                else: filled.discard(group[idx - size])
            if len(filled) == (size // k): 
                total += 1

            return total

        def solve_group(group):
            total = 0
            for size in range(k, len(word) + 1, k):
                if size == 27 * k: break
                total += solve_size(group, size)
            return total

        return sum(solve_group(group) for group in groups)

7155