214

214. Shortest Palindrome (Hard)

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.

Companies:
Microsoft, Apple

Related Topics:
String, Rolling Hash, String Matching, Hash Function

Similar Questions:

• Longest Palindromic Substring (Medium)
• Implement strStr() (Easy)
• Palindrome Pairs (Hard)

Solution 1. Rolling Hash

Find the longest palindrome starting at index 0.

// OJ: https://leetcode.com/problems/shortest-palindrome/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1) ignoring the space taken by the answer
class Solution {
public:
    string shortestPalindrome(string s) {
        unsigned d = 16777619, h = 0, rh = 0, p = 1, maxLen = 0;
        for (int i = 0; i < s.size(); ++i) {
            h = h * d + s[i] - 'a';
            rh += (s[i] - 'a') * p;
            p *= d;
            if (h == rh) maxLen = i + 1;
        }
        return string(rbegin(s), rbegin(s) + s.size() - maxLen) + s;
    }
};

Solution 2. Manacher

// OJ: https://leetcode.com/problems/shortest-palindrome/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    string shortestPalindrome(string s) {
        int N = s.size();
        string t = "^*";
        for (char c : s) {
            t += c;
            t += '*';
        }
        t += '$';
        vector<int> r(t.size(), 1);
        int j = 1, len = 0;
        for (int i = 2; i <= 2 * N; ++i) {
            int cur = j + r[j] > i ? min(r[2 * j - i], j + r[j] - i) : 1;
            while (t[i - cur] == t[i + cur]) ++cur;
            if (i + cur > j + r[j]) j = i;
            r[i] = cur;
            if (i - cur == 0) len = i - 1;
        }
        return string(rbegin(s), rbegin(s) + N - len) + s;
    }
};