<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 makePrefixArray(self, a):
        a_prefix = [0]
        j, i = 0, 1
        while i < len(a):
            if a[j] == a[i]:
                a_prefix.append(j + 1)
                j += 1
                i += 1
            elif j != 0:
                j = a_prefix[j - 1]
            else:
                a_prefix.append(0)
                i += 1
        return a_prefix

    def findIndices(self, s, p, prefix):
        si, pi = 0, 0
        indices = []
        while si < len(s):
            if s[si] == p[pi]:
                si, pi = si + 1, pi + 1
                if pi == len(p):
                    indices.append(si - len(prefix))
                    pi = prefix[pi - 1]
            else:
                if pi == 0:
                    si += 1
                else:
                    pi = prefix[pi - 1]
        return indices

    def beautifulIndices(self, s: str, a: str, b: str, k: int) -> List[int]:
        # three strings and k
        # find substrings containing a and b + starting index
        # for each i index, check if the j index is valid
        a_prefix, b_prefix = self.makePrefixArray(a), self.makePrefixArray(b)
        a_indices, b_indices = self.findIndices(s, a, a_prefix), self.findIndices(
            s, b, b_prefix
        )
        a_index, b_index = 0, 0
        beautiful = []
        while a_index < len(a_indices) and b_index < len(b_indices):
            if abs(a_indices[a_index] - b_indices[b_index]) <= k:
                beautiful.append(a_indices[a_index])
                a_index += 1
            elif a_indices[a_index] >= b_indices[b_index]:
                b_index += 1
            else:
                a_index += 1

        return beautiful
        
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