<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>

from typing import List
def kmp(needle ,haystack):
    lps=[0]*len(needle)
    prevlps,i = 0,1
    indices=[]
    while i< len(needle):
        if needle[i]==needle[prevlps]:
            lps[i]=prevlps+1
            prevlps+=1
            i+=1
        elif prevlps==0:
            lps[i]=0
            i+=1
        else:
            prevlps=lps[prevlps-1]
    
    i=0
    j=0
    while i<len(haystack):
        if haystack[i]==needle[j]:
            i,j = i+1 ,j+1
        else:
            if j==0:
                i+=1
            else:
                j = lps[j-1]
        if j==len(needle):
            indices.append(i-len(needle))
            j=lps[j-1]
    return indices

class Solution:
    def beautifulIndices(self, s: str, a: str, b: str, k: int) -> List[int]:
        p = len(a)  # Window size
        n = len(s)
        
        # Find all starting indices of substring a
        tp1 = kmp(a,s)
        
        # Find all starting indices of substring b
        tp2 = kmp(b,s)
        
        # Helper function to find the closest index in tp2 within range
        def find_closest_indices(arr, target, k):
            low, high = 0, len(arr) - 1
            closest = []
            while low <= high:
                mid = (low + high) // 2
                if arr[mid] < target - k:
                    low = mid + 1
                elif arr[mid] > target + k:
                    high = mid - 1
                else:
                    # Collect all indices in the range
                    left, right = mid, mid
                    return 1
                    
            return 0
        
        ans = []
        for i in tp1:
            if find_closest_indices(tp2, i, k):
                ans.append(i)
        
        return ans



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