Find Beautiful Indices in the Given Array I
<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>
Two Pointers; String; Binary Search; String Matching; Hash Function; Rolling Hash
class Solution {
public:
    vector<int> beautifulIndices(string s, string patternA, string patternB, int k) {
        vector<int> beautifulIndicesA = kmpSearch(s, patternA);
        vector<int> beautifulIndicesB = kmpSearch(s, patternB);

        sort(beautifulIndicesB.begin(), beautifulIndicesB.end());

        vector<int> result;
        for (int indexA : beautifulIndicesA) {
            int left = lower_bound(beautifulIndicesB.begin(), beautifulIndicesB.end(), indexA - k) - beautifulIndicesB.begin();
            int right = lower_bound(beautifulIndicesB.begin(), beautifulIndicesB.end(), indexA + k + patternB.length()) - beautifulIndicesB.begin();

            left = (left >= 0) ? left : -(left + 1);
            right = (right >= 0) ? right : -(right + 1);

            for (int indexB = left; indexB < right; indexB++) {
                if (abs(beautifulIndicesB[indexB] - indexA) <= k) {
                    result.push_back(indexA);
                    break;
                }
            }
        }

        return result;
    }

private:
    vector<int> kmpSearch(string text, string pattern) {
        vector<int> indices;
        vector<int> pi = computePrefixFunction(pattern);

        int q = 0;
        for (int i = 0; i < text.length(); i++) {
            while (q > 0 && pattern[q] != text[i]) {
                q = pi[q - 1];
            }
            if (pattern[q] == text[i]) {
                q++;
            }
            if (q == pattern.length()) {
                indices.push_back(i - q + 1);
                q = pi[q - 1];
            }
        }

        return indices;
    }

    vector<int> computePrefixFunction(string pattern) {
        int m = pattern.length();
        vector<int> pi(m, 0);
        int k = 0;
        for (int q = 1; q < m; q++) {
            while (k > 0 && pattern[k] != pattern[q]) {
                k = pi[k - 1];
            }
            if (pattern[k] == pattern[q]) {
                k++;
            }
            pi[q] = k;
        }
        return pi;
    }
};
