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# Shortest Palindrome LeetCode Solution

## Problem – Shortest Palindrome LeetCode Solution

You are given a string `s`. You can convert `s` to a palindrome by adding characters in front of it.

Return the shortest palindrome you can find by performing this transformation.

Example 1:

``````Input: s = "aacecaaa"
Output: "aaacecaaa"
``````

Example 2:

``````Input: s = "abcd"
Output: "dcbabcd"
``````

Constraints:

• `0 <= s.length <= 5 * 104`
• `s` consists of lowercase English letters only.

## Shortest Palindrome LeetCode Solution in Python

``````def shortestPalindrome(self, s):
r = s[::-1]
for i in range(len(s) + 1):
if s.startswith(r[i:]):
return r[:i] + s
``````

## Shortest Palindrome LeetCode Solution in Java

``````    int j = 0;
for (int i = s.length() - 1; i >= 0; i--) {
if (s.charAt(i) == s.charAt(j)) { j += 1; }
}
if (j == s.length()) { return s; }
String suffix = s.substring(j);
return new StringBuffer(suffix).reverse().toString() + shortestPalindrome(s.substring(0, j)) + suffix;
``````

## Shortest Palindrome LeetCode Solution in C++

``````class Solution {
public:
string shortestPalindrome(string s) {
string rev_s = s;
reverse(rev_s.begin(), rev_s.end());
string l = s + "#" + rev_s;

vector<int> p(l.size(), 0);
for (int i = 1; i < l.size(); i++) {
int j = p[i - 1];
while (j > 0 && l[i] != l[j])
j = p[j - 1];
p[i] = (j += l[i] == l[j]);
}

return rev_s.substr(0, s.size() - p[l.size() - 1]) + s;
}
};
``````
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