<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>

from bisect import bisect_left, bisect_right

class Solution:
    # KMP function to find all occurrences of the pattern in the string
    def kmp_search(self, text: str, pattern: str):
        # KMP Preprocessing: Build the longest prefix which is also suffix (LPS) array
        lps = [0] * len(pattern)
        j = 0  # length of previous longest prefix suffix
        i = 1
        
        # Build the LPS array
        while i < len(pattern):
            if pattern[i] == pattern[j]:
                j += 1
                lps[i] = j
                i += 1
            else:
                if j != 0:
                    j = lps[j - 1]
                else:
                    lps[i] = 0
                    i += 1
        
        # Perform KMP search to find all starting positions of the pattern in the text
        i = 0  # index for text
        j = 0  # index for pattern
        indices = []
        
        while i < len(text):
            if pattern[j] == text[i]:
                i += 1
                j += 1

            if j == len(pattern):
                # Match found, store the starting index
                indices.append(i - j)
                j = lps[j - 1]
            elif i < len(text) and pattern[j] != text[i]:
                if j != 0:
                    j = lps[j - 1]
                else:
                    i += 1
        
        return indices
    
    def beautifulIndices(self, s: str, a: str, b: str, k: int) -> list[int]:
        # Step 1: Find all occurrences of `a` in `s` using KMP
        special_i = self.kmp_search(s, a)
        
        # Step 2: Find all occurrences of `b` in `s` using KMP
        special_j = self.kmp_search(s, b)
        
        # Step 3: For each i, find if there's any j such that |i - j| <= k using binary search
        ans = []
        
        for i in special_i:
            # Find the range of valid `j` values using binary search
            left = bisect_left(special_j, i - k)
            right = bisect_right(special_j, i + k)
            
            # If there's at least one valid `j` in this range, add `i` to result
            if left < right:
                ans.append(i)
        
        return ans

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