<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]:
        def rabin_karp(pattern, s):
            #Rabin-Karp String matching
            base = 256
            mod = 10 ** 9 + 7
            target_hash = 0
            res = []
            if len(s) < len(pattern): return []
            for i in range(len(pattern)):
                target_hash = (target_hash * base + ord(pattern[i])) % mod
            
            curr_hash = 0
            for i in range(len(pattern)):
                curr_hash = (curr_hash * base + ord(s[i])) % mod
            if curr_hash == target_hash: res.append(0)

            high_base_pow = pow(base, len(pattern)-1, mod)

            l = 0
            r = len(pattern) - 1
            while r+1 < len(s):
                curr_hash = (curr_hash - ord(s[l]) * high_base_pow) % mod
                curr_hash = (curr_hash * base + ord(s[r+1])) % mod
                if curr_hash < 0: curr_hash += mod

                # print(s[l:r+1] + " " + str(curr_hash))
    
                if curr_hash == target_hash:
                    res.append(l+1)
                l += 1
                r += 1

            return res

        def binary_search(target, nums, k):
            lo = 0 
            hi = len(nums) - 1

            while lo <= hi:
                mid = (lo + hi) // 2
                condition = abs(target - nums[mid])
                if condition <= k: return True
                elif nums[mid] > target: hi = mid - 1
                else: lo = mid + 1
            return False
            

        A = rabin_karp(a, s)
        B = rabin_karp(b, s)
        # print(A)
        # print(B)

        res = []

        for value in A:
            if binary_search(value, B, k):
                res.append(value)

        return res
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