<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;= 5 * 10<sup>5</sup></code></li>
	<li><code>1 &lt;= a.length, b.length &lt;= 5 * 10<sup>5</sup></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 = set()

        def kmp(s):
            n = len(s)
            lps = [0] * n # Create the LPS array
            for i in range(1, n): # Start from index 1 (base case lps[0] = 0)
                cur = lps[i-1] # Set cur to the value of the previous LPS
                while cur and s[i] != s[cur]: # If there's a mismatch, move cur to a shorter prefix
                    cur = lps[cur-1] # Move back according to the LPS
                match = s[i] == s[cur] 
                lps[i] = cur + match # If the characters match, increment the LPS value
            return lps # Return the LPS array
        
        # KMP is applied to both the pattern a and the string s
        # concatenated with a separator (#). 
        # By using the LPS array, the algorithm efficiently finds 
        # all positions where a and b match in s.
        alps = kmp(a + "#" + s)
        aIndices = [i - (2*len(a)) for i,v in enumerate(alps) if v >= len(a)]
        blps = kmp(b + "#" + s)
        bIndices = [i - (2*len(b)) for i,v in enumerate(blps) if v >= len(b)]

        # 2-pointer
        res = []
        j = 0
        for i in aIndices:
            while j < len(bIndices) and bIndices[j] < i - k:
                j += 1
            if j < len(bIndices) and bIndices[j] <= i + k:
                res.append(i)
        return res
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