<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;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= a.length, b.length &lt;= 10</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]:
        indices_for_a = self.find_substring_incides(a, s)
        indices_for_b = self.find_substring_incides(b, s)

        print(indices_for_a)
        print(indices_for_b)

        res = []
        # O(N^2)
        # for a in indices_for_a:
        #     for b in indices_for_b:
        #         if self.check_difference(a, b, k):
        #             res.append(a)
        #             break


        # Using 2 pointers to optimize to O(N)
        # |j - i| <= k can write this as two inequalities:
        # i - j <= k and j - i <= k
        # i - k <= j and j <= k + 1
        # which results in: i - k <= j <= k + i
        # so this is the condition that j must follow for a given i
        j = 0
        for i in indices_for_a:
            # Move j to the first valid index where indices_for_b[j] >= i - k
            while j < len(indices_for_b) and indices_for_b[j] < i - k:
                j += 1
            
            # Now check if indices_for_b[j] is within the range [i - k, i + k]
            if j < len(indices_for_b) and abs(indices_for_b[j] - i) <= k:
                res.append(i)

        return res
    
    def find_substring_incides(self, child: str, parent: str) -> List[int]:
        indices = []

        for p in range(len(parent)):
            if child == parent[p : p + len(child)]:
                indices.append(p)
        return indices
    
    def check_difference(self, a_index: int, b_index: int, k: int) -> bool:
        return abs(a_index - b_index) <= k
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