<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>

class Solution:
    def beautifulIndices(self, s: str, a: str, b: str, k: int) -> List[int]:
        res = []
        lenA = len(a)
        lenB = len(b)
        lenS = len(s)

        #precompute valid string instead
        indicesA = [i for i in range(lenS - lenA + 1) if s[i: i + lenA] == a]
        indicesB = [i for i in range(lenS - lenB + 1) if s[i: i + lenB] == b]

        #then check for abs difference
        for i in indicesA:
            for j in indicesB:
                if abs(i - j) <= k:
                    res.append(i)
                    break
        return sorted(res)

        # for i in range(len(s) - lenA + 1):
        #     if s[i:(i + lenA)] == a:
        #         for j in range(i, len(s) - lenB + 1):
        #             if s[j:(j + lenB)] == b:
        #                 if abs(j - i) <= k:
        #                     res.append(i)
        #                     break
        # return sorted(res)

        # res = []
        # len_s = len(s)
        # len_a = len(a)
        # len_b = len(b)
        
        # # Collect all start indices of substrings equal to 'a'
        # indices_a = [i for i in range(len_s - len_a + 1) if s[i:i + len_a] == a]
        
        # # Collect all start indices of substrings equal to 'b'
        # indices_b = [j for j in range(len_s - len_b + 1) if s[j:j + len_b] == b]
        
        # # Check each index i in indices_a to see if there is a valid index j in indices_b
        # for i in indices_a:
        #     for j in indices_b:
        #         if abs(j - i) <= k:
        #             res.append(i)
        #             break  # No need to check further once a valid j is found for this i
        
        # return sorted(res)

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