<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:

        unique = set(word)

        if len(unique) == 1:
            if len(word) - k >= 0:
                return len(word) - k + 1
            return 0
        
        def chk_2(left, right):
            for i in range(left + 1, right + 1):
                if abs(ord(word[i]) - ord(word[i - 1])) > 2: return False
            return True
        
        def chk_occur(freq, left, right):
            for c in set(word[left: right + 1]):
                if freq[c] != k: return False
            return True

        def find(window_size: int):
            left = result = 0
            freq = collections.defaultdict(int)

            for right in range(len(word)):
                freq[word[right]] += 1
                while left < right and right - left + 1 > window_size:
                    freq[word[left]] -= 1
                    left += 1
                if right - left + 1 == window_size and freq[word[right]] == k:
                    if chk_2(left, right) and chk_occur(freq, left, right):
                        result += 1
            return result
        
        result = 0
        for i in range(k, len(set(word)) * k + 1, k):

            cnt = find(i)
            print(i, cnt)
            result += cnt
        return result
            
            


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