| comments | difficulty | edit_url | tags | ||||
|---|---|---|---|---|---|---|---|
true |
Hard |
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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 * 104sconsists of lowercase English letters only.
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] + sclass 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;
}
}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;
}
};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
}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)
}
}
}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;
}
}According to the problem description, we need to reverse the string
The time complexity is
class Solution:
def shortestPalindrome(self, s: str) -> str:
t = s + "#" + s[::-1] + "$"
n = len(t)
next = [0] * n
next[0] = -1
i, j = 2, 0
while i < n:
if t[i - 1] == t[j]:
j += 1
next[i] = j
i += 1
elif j:
j = next[j]
else:
next[i] = 0
i += 1
return s[::-1][: -next[-1]] + sclass Solution {
public String shortestPalindrome(String s) {
String rev = new StringBuilder(s).reverse().toString();
char[] t = (s + "#" + rev + "$").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;
}
}class Solution {
public:
string shortestPalindrome(string s) {
string t = s + "#" + string(s.rbegin(), s.rend()) + "$";
int n = t.size();
int next[n];
next[0] = -1;
next[1] = 0;
for (int i = 2, j = 0; i < n;) {
if (t[i - 1] == t[j]) {
next[i++] = ++j;
} else if (j > 0) {
j = next[j];
} else {
next[i++] = 0;
}
}
return string(s.rbegin(), s.rbegin() + s.size() - next[n - 1]) + s;
}
};func shortestPalindrome(s string) string {
t := s + "#" + reverse(s) + "$"
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)
}function shortestPalindrome(s: string): string {
const rev = s.split('').reverse().join('');
const t = s + '#' + rev + '$';
const n = t.length;
const next: number[] = Array(n).fill(0);
next[0] = -1;
for (let i = 2, j = 0; i < n; ) {
if (t[i - 1] === t[j]) {
next[i++] = ++j;
} else if (j > 0) {
j = next[j];
} else {
next[i++] = 0;
}
}
return rev.slice(0, -next[n - 1]) + s;
}public class Solution {
public string ShortestPalindrome(string s) {
char[] t = (s + "#" + new string(s.Reverse().ToArray()) + "$").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;
}
}