<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]:
        # KMP
        def kmp(s1: str, s2: str) -> List[int]:
            n = len(s1)
            m = len(s2)

            lps = [0] * m
            prevLPS, i = 0, 1
            while i < m:
                if s2[i] == s2[prevLPS]:
                    prevLPS = lps[i] = prevLPS + 1
                    i += 1
                elif prevLPS == 0:
                    lps[i] = 0
                    i += 1
                else:
                    prevLPS = lps[prevLPS - 1]
            
            ans = []
            i = j = 0
            while i < n:
                if s1[i] == s2[j]:
                    i += 1
                    j += 1
                elif j == 0:
                    i += 1
                else:
                    j = lps[j - 1]
                
                if j == m:
                    ans.append(i - m)
                    j = lps[j - 1]
            
            return ans
        
        # Finding pairs i == j +- k
        iIndexes = kmp(s, a)
        jIndexes = kmp(s, b)
        ans = []

        i = j = 0
        while i < len(iIndexes):
            if j == len(jIndexes): break
            elif -k <= iIndexes[i] - jIndexes[j] <= k:
                ans.append(iIndexes[i])
                i += 1
            elif iIndexes[i] - jIndexes[j] > k:
                j += 1
            else:
                i += 1
        return ans



        return ans
768