## 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 * 10`^{4}

`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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