<p>You are given a <strong>0-indexed</strong> string <code>s</code>, a string <code>a</code>, a string <code>b</code>, and an integer <code>k</code>.</p>

<p>An index <code>i</code> is <strong>beautiful</strong> if:</p>

<ul>
	<li><code>0 &lt;= i &lt;= s.length - a.length</code></li>
	<li><code>s[i..(i + a.length - 1)] == a</code></li>
	<li>There exists an index <code>j</code> such that:
	<ul>
		<li><code>0 &lt;= j &lt;= s.length - b.length</code></li>
		<li><code>s[j..(j + b.length - 1)] == b</code></li>
		<li><code>|j - i| &lt;= k</code></li>
	</ul>
	</li>
</ul>

<p>Return <em>the array that contains beautiful indices in <strong>sorted order from smallest to largest</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;isawsquirrelnearmysquirrelhouseohmy&quot;, a = &quot;my&quot;, b = &quot;squirrel&quot;, k = 15
<strong>Output:</strong> [16,33]
<strong>Explanation:</strong> There are 2 beautiful indices: [16,33].
- The index 16 is beautiful as s[16..17] == &quot;my&quot; and there exists an index 4 with s[4..11] == &quot;squirrel&quot; and |16 - 4| &lt;= 15.
- The index 33 is beautiful as s[33..34] == &quot;my&quot; and there exists an index 18 with s[18..25] == &quot;squirrel&quot; and |33 - 18| &lt;= 15.
Thus we return [16,33] as the result.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcd&quot;, a = &quot;a&quot;, b = &quot;a&quot;, k = 4
<strong>Output:</strong> [0]
<strong>Explanation:</strong> There is 1 beautiful index: [0].
- The index 0 is beautiful as s[0..0] == &quot;a&quot; and there exists an index 0 with s[0..0] == &quot;a&quot; and |0 - 0| &lt;= 4.
Thus we return [0] as the result.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= a.length, b.length &lt;= 10</code></li>
	<li><code>s</code>, <code>a</code>, and <code>b</code> contain only lowercase English letters.</li>
</ul>

from collections import Counter, defaultdict
class Solution:

    def findIdx(self, a, s):
        ahash = Counter(a)
        needed = len(ahash.values())
        afound = []

        window = defaultdict(int)
        for i in range(min(len(a), len(s))):
            val = s[i]
            window[val] += 1

            #check if added a matched or exceeded the necessary amount
            if ahash[val] == window[val]:
                needed -= 1
            elif ahash[val] == window[val] - 1:
                needed += 1
            
            if needed == 0 and a == s[:len(a)]:
                #the first characters len(a) characters in S match A!
                afound.append(0)

        if len(a)>=len(s):
            return afound 
    

        for r in range(len(a), len(s)):
            i = r - len(a) 
            #need to add r and remove i

            #adding r
            val = s[r]
            window[val] += 1
            
            #check if added a matched or exceeded the necessary amount
            if ahash[val] == window[val]:
                needed -= 1
            elif ahash[val] == window[val] - 1:
                needed += 1

            #removing i, check if removed got a match or if it went under
            val = s[i]
            window[val] -= 1

            #check if added a matched or exceeded the necessary amount
            if ahash[val] == window[val]:
                needed -= 1
            elif ahash[val] == window[val] + 1:
                needed += 1
            
            if needed == 0 and a == s[i+1:r+1]:
                # print("comapring", a)
                # print("found", s[i+1:r+1])
                afound.append(i+1)
        return afound

    def beautifulIndices(self, s: str, a: str, b: str, k: int) -> List[int]:

        afound = self.findIdx(a, s)
        bfound = self.findIdx(b, s)
        print("afound", afound)
        print("bfound", bfound)

        ans = []
        for i in afound:

            for j in bfound:
                if abs(j-i) <= k:
                    ans.append(i)
                    break



        return ans



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