<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]:
        n = len(s)
        set_b = []
        set_a = []
        ans = []

        def get_lps(p):
            m = len(p)
            lps = [0] * m

            prev = 0
            i = 1
            while i < m:
                if p[i] == p[prev]:
                    prev += 1
                    lps[i] = prev
                    i += 1
                else:
                    if prev == 0:
                        lps[i] = 0
                        i += 1
                    else:
                        prev = lps[prev - 1]

            return lps


        lps = get_lps(b)
        # print(lps)

        # search for occurrences of b
        i = 0
        j = 0
        while i < n:
            if s[i] == b[j]:
                j += 1
                i += 1
            else:
                if j == 0:
                    i += 1
                else:
                    j = lps[j - 1]
            

            if j == len(b):
                set_b.append(i - len(b))
                j = lps[j - 1]


        lps = get_lps(a)
        
        # search for occurrences of a
        i = 0
        j = 0
        while i < n:
            if s[i] == a[j]:
                j += 1
                i += 1
            else:
                if j == 0:
                    i += 1
                else:
                    j = lps[j - 1]
            

            if j == len(a):
                set_a.append(i - len(a))
                j = lps[j - 1]



        for i in range(len(set_a)):
    
            
            idx_a = set_a[i]
        
            
            b_idx_left = bisect_left(set_b, idx_a - k)
            b_idx_right = bisect_right(set_b, idx_a + k)
            
            # check if there's any index in set_b that is within k distance from i
            if b_idx_right - b_idx_left >= 1:
                ans.append(idx_a)
                
                
        return ans
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