<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]:
        base = 26
        mod = 100000000069

        def get_hash(s, W):
            hash = 0
            for i in range(W):
                hash = (hash * base + ord(s[i])) % mod
            return hash

        def rolling_hash(s, W):
            n = len(s)
            power = [1] * (n + 1)
            for i in range(1, n + 1):
                power[i] = (power[i - 1] * base) % mod
            
            res = [0] * (n - W + 1)

            curr_hash = get_hash(s, W)
            
            # roll for haystack
            res[0] = curr_hash
            for i in range(1, n - W + 1):
                curr_hash = (curr_hash - power[W - 1] * ord(s[i - 1])) % mod
                curr_hash = (curr_hash * base + ord(s[i + W - 1])) % mod
                res[i] = curr_hash

            return res

        if len(a) > len(s) or len(b) > len(s): return []
        hsa, hsb = rolling_hash(s, len(a)), rolling_hash(s, len(b))
        ha, hb = get_hash(a, len(a)), get_hash(b, len(b))
        a, b = [], []
        for id, h in enumerate(hsa):
            if h == ha: a.append(id)
        for id, h in enumerate(hsb):
            if h == hb: b.append(id)

        res = []
        for i in a:
            pos = bisect.bisect_left(b, i)
            if 0 <= pos < len(b) and abs(i - b[pos]) <= k: res.append(i);continue
            if pos - 1 >= 0 and abs(i - b[pos-1]) <= k: res.append(i);continue
            if pos + 1 < len(b) and abs(i - b[pos+1]) <= k: res.append(i);continue
        return res
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