<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:
        uniq=set(word)
        if len(uniq)==1:
            if k<=len(word):
                return len(word)-k+1
            else:
                return 0
        def adj(word,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(word,left,right,freq,k):
            seen=set(word[left:right+1])
            for ch in seen:
                if freq[ch]!=k:
                    return False
            return True
        def finder(word, window_len, k):
            left, right, ans = 0 , 0, 0
            freq = defaultdict(int)
            
            while right < len(word):
                freq[word[right]] += 1
                
                while right -  left + 1 > window_len:
                    freq[word[left]] -= 1
                    left += 1
                
                if freq[word[right]] == k and (right - left + 1) == window_len:
                    if adj(word, left, right) and chk(word, left, right, freq, k):
                        ans += 1
                right += 1
            return ans   
        ans = 0
        for i in range(1*k , (len(set(word)) * k) + 1, k):
            ans += finder(word, i, k)  
        return ans
8076