Find Beautiful Indices in the Given Array II
<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>
Two Pointers; String; Binary Search; String Matching; Hash Function; Rolling Hash
class Solution:
    def beautifulIndices(self, s: str, a: str, b: str, k: int) -> List[int]:
        def build_prefix_function(pattern):
            prefix_function = [0] * len(pattern)
            j = 0
            for i in range(1, len(pattern)):
                while j > 0 and pattern[i] != pattern[j]:
                    j = prefix_function[j - 1]
                if pattern[i] == pattern[j]:
                    j += 1
                prefix_function[i] = j
            return prefix_function

        def kmp_search(pattern, text, prefix_function):
            occurrences = []
            j = 0
            for i in range(len(text)):
                while j > 0 and text[i] != pattern[j]:
                    j = prefix_function[j - 1]
                if text[i] == pattern[j]:
                    j += 1
                if j == len(pattern):
                    occurrences.append(i - j + 1)
                    j = prefix_function[j - 1]
            return occurrences

        prefix_a = build_prefix_function(a)
        prefix_b = build_prefix_function(b)

        resa = kmp_search(a, s, prefix_a)
        resb = kmp_search(b, s, prefix_b)

        res = []
        print(resa, resb)
        i = 0
        j = 0
        while i < len(resa):
            while j < len(resb):
                if abs(resb[j] - resa[i]) <= k:
                    res.append(resa[i])
                    break
                elif j + 1 < len(resb) and abs(resb[j + 1] - resa[i]) < abs(
                    resb[j] - resa[i]
                ):
                    j += 1
                else:
                    break
            i += 1
        return res
