<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:
        # window cannot be more than 26*k
        # window sie of k, 2k, 3k, 4k..... 26k

        n = len(word)

        def check(curr):
            for char in "abcdefghijklmnopqrstuvwxyz":
                if char in curr and curr[char] != k:
                    return 0
            return 1

        
        def window(i):
            
            curr = {}
            right, s = -1, i*k
            
            res = 0
            for left in range(n-s+1):

                while right - left + 1 < s and (right < left or abs(ord(word[right+1]) - ord(word[right])) <= 2):
                    right += 1
                    if word[right] not in curr:
                        curr[word[right]] = 0
                    curr[word[right]] += 1

                if right - left + 1 == s:
                    res += check(curr)

                if left > right:
                    right = left
                else:
                    curr[word[left]] -= 1
                    if curr[word[left]] == 0:
                        curr.pop(word[left])
            return res


        count = 0

        for i in range(1, min(26, n // k) + 1):

            count += window(i)

        return count

5389