<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]:
        if len(s) < len(a) or len(s) < len(b):
            return []
        
        MOD = 10**9 + 7
        BASE = 31

        def rolling_hash(st) -> List[int]:
            m, n = len(st), len(s)
            powers = [1] * (n + 1)
            indices = []

            for i in range(1, n+1):
                powers[i] = (powers[i - 1] * BASE) % MOD
        
            curr_hash = 0
            for i in range(m):
                curr_hash = (curr_hash * BASE + ord(s[i])) % MOD
            
            hash_st = 0
            for i in range(m):
                hash_st = (hash_st * BASE + ord(st[i])) % MOD
            
            if curr_hash == hash_st:
                indices.append(0)
            
            for i in range(1, n - m + 1):
                curr_hash = (curr_hash - powers[m - 1] * ord(s[i - 1])) % MOD
                curr_hash = (curr_hash * BASE + ord(s[i + m - 1])) % MOD
                if curr_hash == hash_st:
                    indices.append(i)
            return indices

        indices_a = rolling_hash(a)
        indices_b = rolling_hash(b)

        if not indices_b or not indices_a:
            return []

        res = []
        for ind_a in indices_a:
            # (j >= i - k)
            lo, hi = 0, len(indices_b) - 1
            found = False
            while lo <= hi:
                mid = lo + (hi - lo) // 2
                if (ind_a - k) <= indices_b[mid] <= (ind_a + k):
                    found = True
                    break
                elif indices_b[mid] <= (ind_a - k):
                    lo = mid + 1
                elif indices_b[mid] > (ind_a + k):
                    hi = mid - 1
            if found:
                res.append(ind_a)
        return res
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