<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>

def check(dic,k):
    L=[]
    for x in dic:
        if dic[x]!=0:
            L.append(dic[x])
    if len(set(L))==1 and L[0]==k:
        return True
    return False
def compute(L,size,k):
    dic={}
    count=0
    l,r=0,0
    ans=0
    while r<len(L):
        if count==0:
            dic[L[r]]=dic.get(L[r],0)+1
            r=r+1
            count=count+1
        else:
            if abs((ord(L[r])-ord(L[r-1])))<=2:
                dic[L[r]]=dic.get(L[r],0)+1
                count=count+1
                r=r+1
            else:
                dic={}
                l=r
                count=0
                continue
        if count>size:
            dic[L[l]]=dic[L[l]]-1
            l=l+1
            count=count-1
        if count==size:
            if check(dic,k)==True:
                ans=ans+1
    return ans

class Solution:
    def countCompleteSubstrings(self, L: str, k: int) -> int:
        num=len(set(L))
        ans=0
        for size in range (k,min(len(L)+1,(k*num)+1),k):
            if size==1:
                ans=ans+len(L)
                continue
            count=compute(L,size,k)
            # print(count,size)
            ans=ans+count
        return ans

        
3471