<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 is_same_frequency(self, freq, k: int):
    return all([freq[i] == k or freq[i] == 0 for i in freq])
  def count_substrings(self, s: str, k: int) -> int:
    count, distinct = 0, len(set([i for i in s]))
    for l in range(1, distinct + 1):
      max_l, freq = l * k, defaultdict(int)
      pos_s, pos_e = 0, max_l - 1
      for i in range(pos_s, min(pos_e + 1, len(s))):
        freq[s[i]] += 1
      while pos_e < len(s):
        if self.is_same_frequency(freq, k):
          count += 1
        freq[s[pos_s]] -= 1
        pos_s, pos_e = pos_s + 1, pos_e + 1
        if pos_e < len(s):
          freq[s[pos_e]] += 1
    return count
  def countCompleteSubstrings(self, word: str, k: int) -> int:
    groups = [[]]
    for c in word:
      if len(groups[-1]) == 0:
        groups[-1].append(c)
      elif abs(ord(groups[-1][-1]) - ord(c)) <= 2:
        groups[-1].append(c)
      else:
        groups.append([c])
    res = 0
    for g in groups:
      res += self.count_substrings(''.join(g), k)
    return res
4085