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214. Shortest Palindrome

Description

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.

Solutions

Solution 1

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class Solution:
    def shortestPalindrome(self, s: str) -> str:
        base = 131
        mod = 10**9 + 7
        n = len(s)
        prefix = suffix = 0
        mul = 1
        idx = 0
        for i, c in enumerate(s):
            prefix = (prefix * base + (ord(c) - ord('a') + 1)) % mod
            suffix = (suffix + (ord(c) - ord('a') + 1) * mul) % mod
            mul = (mul * base) % mod
            if prefix == suffix:
                idx = i + 1
        return s if idx == n else s[idx:][::-1] + s

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class Solution {
    public String shortestPalindrome(String s) {
        int base = 131;
        int mul = 1;
        int mod = (int) 1e9 + 7;
        int prefix = 0, suffix = 0;
        int idx = 0;
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            int t = s.charAt(i) - 'a' + 1;
            prefix = (int) (((long) prefix * base + t) % mod);
            suffix = (int) ((suffix + (long) t * mul) % mod);
            mul = (int) (((long) mul * base) % mod);
            if (prefix == suffix) {
                idx = i + 1;
            }
        }
        if (idx == n) {
            return s;
        }
        return new StringBuilder(s.substring(idx)).reverse().toString() + s;
    }
}

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typedef unsigned long long ull;

class Solution {
public:
    string shortestPalindrome(string s) {
        int base = 131;
        ull mul = 1;
        ull prefix = 0;
        ull suffix = 0;
        int idx = 0, n = s.size();
        for (int i = 0; i < n; ++i) {
            int t = s[i] - 'a' + 1;
            prefix = prefix * base + t;
            suffix = suffix + mul * t;
            mul *= base;
            if (prefix == suffix) idx = i + 1;
        }
        if (idx == n) return s;
        string x = s.substr(idx, n - idx);
        reverse(x.begin(), x.end());
        return x + s;
    }
};

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func shortestPalindrome(s string) string {
    n := len(s)
    base, mod := 131, int(1e9)+7
    prefix, suffix, mul := 0, 0, 1
    idx := 0
    for i, c := range s {
        t := int(c-'a') + 1
        prefix = (prefix*base + t) % mod
        suffix = (suffix + t*mul) % mod
        mul = (mul * base) % mod
        if prefix == suffix {
            idx = i + 1
        }
    }
    if idx == n {
        return s
    }
    x := []byte(s[idx:])
    for i, j := 0, len(x)-1; i < j; i, j = i+1, j-1 {
        x[i], x[j] = x[j], x[i]
    }
    return string(x) + s
}

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impl Solution {
    pub fn shortest_palindrome(s: String) -> String {
        let base = 131;
        let (mut idx, mut prefix, mut suffix, mut mul) = (0, 0, 0, 1);
        for (i, c) in s.chars().enumerate() {
            let t = (c as u64) - ('0' as u64) + 1;
            prefix = prefix * base + t;
            suffix = suffix + t * mul;
            mul *= base;
            if prefix == suffix {
                idx = i + 1;
            }
        }
        if idx == s.len() {
            s
        } else {
            let x: String = (&s[idx..]).chars().rev().collect();
            String::from(x + &s)
        }
    }
}

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public class Solution {
    public string ShortestPalindrome(string s) {
        int baseValue = 131;
        int mul = 1;
        int mod = (int)1e9 + 7;
        int prefix = 0, suffix = 0;
        int idx = 0;
        int n = s.Length;

        for (int i = 0; i < n; ++i) {
            int t = s[i] - 'a' + 1;
            prefix = (int)(((long)prefix * baseValue + t) % mod);
            suffix = (int)((suffix + (long)t * mul) % mod);
            mul = (int)(((long)mul * baseValue) % mod);
            if (prefix == suffix) {
                idx = i + 1;
            }
        }

        if (idx == n) {
            return s;
        }

        return new string(s.Substring(idx).Reverse().ToArray()) + s;
    }
}

Solution 2: KMP Algorithm

According to the problem description, we need to reverse the string $s$ to obtain the string $\textit{rev}$, and then find the longest common part of the suffix of the string $\textit{rev}$ and the prefix of the string $s$. We can use the KMP algorithm to concatenate the string $s$ and the string $\textit{rev}$ and find the longest common part of the longest prefix and the longest suffix.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.

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class Solution:
    def shortestPalindrome(self, s: str) -> str:
        t = s + "#" + s[::-1] + "$"
        n = len(t)
        next = [0] * n
        next[0] = -1
        i, j = 2, 0
        while i < n:
            if t[i - 1] == t[j]:
                j += 1
                next[i] = j
                i += 1
            elif j:
                j = next[j]
            else:
                next[i] = 0
                i += 1
        return s[::-1][: -next[-1]] + s

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class Solution {
    public String shortestPalindrome(String s) {
        String rev = new StringBuilder(s).reverse().toString();
        char[] t = (s + "#" + rev + "$").toCharArray();
        int n = t.length;
        int[] next = new int[n];
        next[0] = -1;
        for (int i = 2, j = 0; i < n;) {
            if (t[i - 1] == t[j]) {
                next[i++] = ++j;
            } else if (j > 0) {
                j = next[j];
            } else {
                next[i++] = 0;
            }
        }
        return rev.substring(0, s.length() - next[n - 1]) + s;
    }
}

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class Solution {
public:
    string shortestPalindrome(string s) {
        string t = s + "#" + string(s.rbegin(), s.rend()) + "$";
        int n = t.size();
        int next[n];
        next[0] = -1;
        next[1] = 0;
        for (int i = 2, j = 0; i < n;) {
            if (t[i - 1] == t[j]) {
                next[i++] = ++j;
            } else if (j > 0) {
                j = next[j];
            } else {
                next[i++] = 0;
            }
        }
        return string(s.rbegin(), s.rbegin() + s.size() - next[n - 1]) + s;
    }
};

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func shortestPalindrome(s string) string {
    t := s + "#" + reverse(s) + "$"
    n := len(t)
    next := make([]int, n)
    next[0] = -1
    for i, j := 2, 0; i < n; {
        if t[i-1] == t[j] {
            j++
            next[i] = j
            i++
        } else if j > 0 {
            j = next[j]
        } else {
            next[i] = 0
            i++
        }
    }
    return reverse(s)[:len(s)-next[n-1]] + s
}

func reverse(s string) string {
    t := []byte(s)
    for i, j := 0, len(t)-1; i < j; i, j = i+1, j-1 {
        t[i], t[j] = t[j], t[i]
    }
    return string(t)
}

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function shortestPalindrome(s: string): string {
    const rev = s.split('').reverse().join('');
    const t = s + '#' + rev + '$';
    const n = t.length;
    const next: number[] = Array(n).fill(0);
    next[0] = -1;
    for (let i = 2, j = 0; i < n; ) {
        if (t[i - 1] === t[j]) {
            next[i++] = ++j;
        } else if (j > 0) {
            j = next[j];
        } else {
            next[i++] = 0;
        }
    }
    return rev.slice(0, -next[n - 1]) + s;
}

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public class Solution {
    public string ShortestPalindrome(string s) {
        char[] t = (s + "#" + new string(s.Reverse().ToArray()) + "$").ToCharArray();
        int n = t.Length;
        int[] next = new int[n];
        next[0] = -1;
        for (int i = 2, j = 0; i < n;) {
            if (t[i - 1] == t[j]) {
                next[i++] = ++j;
            } else if (j > 0) {
                j = next[j];
            } else {
                next[i++] = 0;
            }
        }
        return new string(s.Substring(next[n - 1]).Reverse().ToArray()).Substring(0, s.Length - next[n - 1]) + s;
    }
}

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