<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:
    bool checkPal(string& s, int b, int e){
        if (b >= e) return true;
        if (s[b] != s[e]) return false;
        return checkPal(s, b + 1, e - 1);
    }


    string shortestPalindrome(string s) {
        int n = s.size();
        if (n <= 1) return s;
        char fr = s.front();
        unordered_map<char, bool> temp;
        stack<int> possibleIndex;
        for (int i = 0; i < n; i++){
            if (temp.contains(s[i])){
                temp.erase(s[i]);
            } else {
                temp.insert({s[i], true});
            }
            if (s[i] == fr && temp.size() < 2){
                possibleIndex.push(i);
            }
        }

        while (!possibleIndex.empty()){
            if (checkPal(s, 0, possibleIndex.top())){
                break;
            } else {
                possibleIndex.pop();
            }
        }

        int b = possibleIndex.top();
        string res = s.substr(possibleIndex.top() + 1, s.size() - possibleIndex.top());
        reverse(begin(res), end(res));
        res = res + s;
        return res;
    }
};
14500