<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:
        l, res = 0, 0

        for r in range(1, len(word) + 1):
            if r == len(word) or abs(ord(word[r]) - ord(word[r - 1])) > 2:
                res += self.solve(word[l:r], k)
                l = r
        return res

    def solve(self, word, k):
        #print(word)
        result = 0
        for i in range(k, min(26 * k + 1, len(word) + 1), k):
            #print(str(i))
            counter = Counter()
            l = 0
            for r in range(len(word)):
                found = False
                counter[word[r]] += 1
                #print('Counter: ' + str(counter) + '[r, l]: [' + str(r) + ', ' + str(l) + '], r - l + 1 : ' + str(r - l + 1) + ', k : ' + str(k))
                if r - l + 1 == i:                                        
                    for c in counter:
                        if counter[c] and counter[c] != k:
                            found = True
                            break
                    if not found:
                        result += 1
                    
                    counter[word[l]] -= 1
                    l += 1
        return result


3164