3504. Longest Palindrome After Substring Concatenation II
Description
You are given two strings, s and t.
You can create a new string by selecting a substring from s (possibly empty) and a substring from t (possibly empty), then concatenating them in order.
Return the length of the longest palindrome that can be formed this way.
Example 1:
Input: s = "a", t = "a"
Output: 2
Explanation:
Concatenating "a" from s and "a" from t results in "aa", which is a palindrome of length 2.
Example 2:
Input: s = "abc", t = "def"
Output: 1
Explanation:
Since all characters are different, the longest palindrome is any single character, so the answer is 1.
Example 3:
Input: s = "b", t = "aaaa"
Output: 4
Explanation:
Selecting "aaaa" from t is the longest palindrome, so the answer is 4.
Example 4:
Input: s = "abcde", t = "ecdba"
Output: 5
Explanation:
Concatenating "abc" from s and "ba" from t results in "abcba", which is a palindrome of length 5.
Constraints:
1 <= s.length, t.length <= 1000sandtconsist of lowercase English letters.
Solutions
Solution 1: Enumerate Palindrome Centers + Dynamic Programming
According to the problem description, the concatenated palindrome string can be composed entirely of string $s$, entirely of string $t$, or a combination of both strings $s$ and $t$. Additionally, there may be extra palindromic substrings in either string $s$ or $t$.
Therefore, we first reverse string $t$ and preprocess arrays $\textit{g1}$ and $\textit{g2}$, where $\textit{g1}[i]$ represents the length of the longest palindromic substring starting at index $i$ in string $s$, and $\textit{g2}[i]$ represents the length of the longest palindromic substring starting at index $i$ in string $t$.
We can initialize the answer $\textit{ans}$ as the maximum value in $\textit{g1}$ and $\textit{g2}$.
Next, we define $\textit{f}[i][j]$ as the length of the palindromic substring ending at the $i$-th character of string $s$ and the $j$-th character of string $t$.
For $\textit{f}[i][j]$, if $s[i - 1]$ equals $t[j - 1]$, then $\textit{f}[i][j] = \textit{f}[i - 1][j - 1] + 1$. We then update the answer:
$$ \textit{ans} = \max(\textit{ans}, \textit{f}[i][j] \times 2 + (0 \text{ if } i \geq m \text{ else } \textit{g1}[i])) \ \textit{ans} = \max(\textit{ans}, \textit{f}[i][j] \times 2 + (0 \text{ if } j \geq n \text{ else } \textit{g2}[j])) $$
Finally, we return the answer $\textit{ans}$.
The time complexity is $O(m \times (m + n))$, and the space complexity is $O(m \times n)$, where $m$ and $n$ are the lengths of strings $s$ and $t$, respectively.
Python3
class Solution:
def longestPalindrome(self, s: str, t: str) -> int:
def expand(s: str, g: List[int], l: int, r: int):
while l >= 0 and r < len(s) and s[l] == s[r]:
g[l] = max(g[l], r - l + 1)
l, r = l - 1, r + 1
def calc(s: str) -> List[int]:
n = len(s)
g = [0] * n
for i in range(n):
expand(s, g, i, i)
expand(s, g, i, i + 1)
return g
m, n = len(s), len(t)
t = t[::-1]
g1, g2 = calc(s), calc(t)
ans = max(*g1, *g2)
f = [[0] * (n + 1) for _ in range(m + 1)]
for i, a in enumerate(s, 1):
for j, b in enumerate(t, 1):
if a == b:
f[i][j] = f[i - 1][j - 1] + 1
ans = max(ans, f[i][j] * 2 + (0 if i >= m else g1[i]))
ans = max(ans, f[i][j] * 2 + (0 if j >= n else g2[j]))
return ans
Java
class Solution {
public int longestPalindrome(String S, String T) {
char[] s = S.toCharArray();
char[] t = new StringBuilder(T).reverse().toString().toCharArray();
int m = s.length, n = t.length;
int[] g1 = calc(s), g2 = calc(t);
int ans = Math.max(Arrays.stream(g1).max().getAsInt(), Arrays.stream(g2).max().getAsInt());
int[][] f = new int[m + 1][n + 1];
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (s[i - 1] == t[j - 1]) {
f[i][j] = f[i - 1][j - 1] + 1;
ans = Math.max(ans, f[i][j] * 2 + (i < m ? g1[i] : 0));
ans = Math.max(ans, f[i][j] * 2 + (j < n ? g2[j] : 0));
}
}
}
return ans;
}
private void expand(char[] s, int[] g, int l, int r) {
while (l >= 0 && r < s.length && s[l] == s[r]) {
g[l] = Math.max(g[l], r - l + 1);
--l;
++r;
}
}
private int[] calc(char[] s) {
int n = s.length;
int[] g = new int[n];
for (int i = 0; i < n; ++i) {
expand(s, g, i, i);
expand(s, g, i, i + 1);
}
return g;
}
}
C++
class Solution {
public:
int longestPalindrome(string s, string t) {
int m = s.size(), n = t.size();
ranges::reverse(t);
vector<int> g1 = calc(s), g2 = calc(t);
int ans = max(ranges::max(g1), ranges::max(g2));
vector<vector<int>> f(m + 1, vector<int>(n + 1));
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (s[i - 1] == t[j - 1]) {
f[i][j] = f[i - 1][j - 1] + 1;
ans = max(ans, f[i][j] * 2 + (i < m ? g1[i] : 0));
ans = max(ans, f[i][j] * 2 + (j < n ? g2[j] : 0));
}
}
}
return ans;
}
private:
void expand(const string& s, vector<int>& g, int l, int r) {
while (l >= 0 && r < s.size() && s[l] == s[r]) {
g[l] = max(g[l], r - l + 1);
--l;
++r;
}
}
vector<int> calc(const string& s) {
int n = s.size();
vector<int> g(n, 0);
for (int i = 0; i < n; ++i) {
expand(s, g, i, i);
expand(s, g, i, i + 1);
}
return g;
}
};
Go
func longestPalindrome(s, t string) int {
m, n := len(s), len(t)
t = reverse(t)
g1, g2 := calc(s), calc(t)
ans := max(slices.Max(g1), slices.Max(g2))
f := make([][]int, m+1)
for i := range f {
f[i] = make([]int, n+1)
}
for i := 1; i <= m; i++ {
for j := 1; j <= n; j++ {
if s[i-1] == t[j-1] {
f[i][j] = f[i-1][j-1] + 1
a, b := 0, 0
if i < m {
a = g1[i]
}
if j < n {
b = g2[j]
}
ans = max(ans, f[i][j]*2+a)
ans = max(ans, f[i][j]*2+b)
}
}
}
return ans
}
func calc(s string) []int {
n, g := len(s), make([]int, len(s))
for i := 0; i < n; i++ {
expand(s, g, i, i)
expand(s, g, i, i+1)
}
return g
}
func expand(s string, g []int, l, r int) {
for l >= 0 && r < len(s) && s[l] == s[r] {
g[l] = max(g[l], r-l+1)
l, r = l-1, r+1
}
}
func reverse(s string) string {
r := []rune(s)
slices.Reverse(r)
return string(r)
}
TypeScript
function longestPalindrome(s: string, t: string): number {
function expand(s: string, g: number[], l: number, r: number): void {
while (l >= 0 && r < s.length && s[l] === s[r]) {
g[l] = Math.max(g[l], r - l + 1);
l--;
r++;
}
}
function calc(s: string): number[] {
const n = s.length;
const g: number[] = Array(n).fill(0);
for (let i = 0; i < n; i++) {
expand(s, g, i, i);
expand(s, g, i, i + 1);
}
return g;
}
const m = s.length,
n = t.length;
t = t.split('').reverse().join('');
const g1 = calc(s);
const g2 = calc(t);
let ans = Math.max(...g1, ...g2);
const f: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
for (let i = 1; i <= m; i++) {
for (let j = 1; j <= n; j++) {
if (s[i - 1] === t[j - 1]) {
f[i][j] = f[i - 1][j - 1] + 1;
ans = Math.max(ans, f[i][j] * 2 + (i >= m ? 0 : g1[i]));
ans = Math.max(ans, f[i][j] * 2 + (j >= n ? 0 : g2[j]));
}
}
}
return ans;
}