<p>You are given a string <code>s</code>. You can convert <code>s</code> to a <span data-keyword="palindrome-string">palindrome</span> by adding characters in front of it.</p>

<p>Return <em>the shortest palindrome you can find by performing this transformation</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "aacecaaa"
<strong>Output:</strong> "aaacecaaa"
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "abcd"
<strong>Output:</strong> "dcbabcd"
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= s.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>s</code> consists of lowercase English letters only.</li>
</ul>

class Solution {
public:
    string shortestPalindrome(string s) {
        if (s.empty()) return s;
        
        string rev_s = s;
        reverse(rev_s.begin(), rev_s.end());
        string combined = s + "#" + rev_s;
        
        vector<int> z = calculateZ(combined);
        
        int max_prefix = 0;
        for (int i = s.length() + 1; i < z.size(); ++i) {
            if (z[i] == combined.length() - i) {
                max_prefix = z[i];
                break;
            }
        }
        
        return rev_s.substr(0, s.length() - max_prefix) + s;
    }
    
private:
    vector<int> calculateZ(const string& s) {
        int n = s.length();
        vector<int> z(n, 0);
        int l = 0, r = 0;
        
        for (int i = 1; i < n; ++i) {
            if (i <= r) {
                z[i] = min(r - i + 1, z[i - l]);
            }
            while (i + z[i] < n && s[z[i]] == s[i + z[i]]) {
                ++z[i];
            }
            if (i + z[i] - 1 > r) {
                l = i;
                r = i + z[i] - 1;
            }
        }
        
        return z;
    }
};
//KDS
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