# 5. Longest Palindromic Substring

## Description

Given a string s, return the longest palindromic substring in s.

Example 1:

Output: "bab"
Explanation: "aba" is also a valid answer.

Example 2:

Input: s = "cbbd"
Output: "bb"

Constraints:

• 1 <= s.length <= 1000
• s consist of only digits and English letters.

## Solutions

Solution 1: Dynamic Programming

We define $f[i][j]$ to represent whether the string $s[i..j]$ is a palindrome, initially $f[i][j] = true$.

Next, we define variables $k$ and $mx$, where $k$ represents the starting position of the longest palindrome, and $mx$ represents the length of the longest palindrome. Initially, $k = 0$, $mx = 1$.

Considering $f[i][j]$, if $s[i] = s[j]$, then $f[i][j] = f[i + 1][j - 1]$; otherwise, $f[i][j] = false$. If $f[i][j] = true$ and $mx < j - i + 1$, then we update $k = i$, $mx = j - i + 1$.

Since $f[i][j]$ depends on $f[i + 1][j - 1]$, we need to ensure that $i + 1$ is before $j - 1$, so we need to enumerate $i$ from large to small, and enumerate $j$ from small to large.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the length of the string $s$.

Solution 2: Enumerate Palindrome Midpoint

We can enumerate the midpoint of the palindrome, spread to both sides, and find the longest palindrome.

The time complexity is $O(n^2)$, and the space complexity is $O(1)$. Here, $n$ is the length of the string $s$.

• class Solution {
public String longestPalindrome(String s) {
int n = s.length();
boolean[][] f = new boolean[n][n];
for (var g : f) {
Arrays.fill(g, true);
}
int k = 0, mx = 1;
for (int i = n - 2; i >= 0; --i) {
for (int j = i + 1; j < n; ++j) {
f[i][j] = false;
if (s.charAt(i) == s.charAt(j)) {
f[i][j] = f[i + 1][j - 1];
if (f[i][j] && mx < j - i + 1) {
mx = j - i + 1;
k = i;
}
}
}
}
return s.substring(k, k + mx);
}
}

• class Solution {
public:
string longestPalindrome(string s) {
int n = s.size();
vector<vector<bool>> f(n, vector<bool>(n, true));
int k = 0, mx = 1;
for (int i = n - 2; ~i; --i) {
for (int j = i + 1; j < n; ++j) {
f[i][j] = false;
if (s[i] == s[j]) {
f[i][j] = f[i + 1][j - 1];
if (f[i][j] && mx < j - i + 1) {
mx = j - i + 1;
k = i;
}
}
}
}
return s.substr(k, mx);
}
};

• class Solution:
def longestPalindrome(self, s: str) -> str:
mlen = 0
start = end = 0
n = len(s)
dp = [ [False] * n for i in range(n) ]

for j in range(n):
for i in range(j + 1):
dp[i][j] = (i == j) or (s[i] == s[j] and j - i == 1) or (s[i] == s[j] and dp[i + 1][j - 1])
if dp[i][j] is True and j - i + 1 > mlen:
mlen = j - i + 1
start = i
end = j

return s[start: end + 1]

######

class Solution:
def longestPalindrome(self, s: str) -> str:
n = len(s)
f = [[True] * n for _ in range(n)]
k, mx = 0, 1
for i in range(n - 2, -1, -1):
for j in range(i + 1, n):
f[i][j] = False
if s[i] == s[j]:
f[i][j] = f[i + 1][j - 1]
if f[i][j] and mx < j - i + 1:
k, mx = i, j - i + 1
return s[k : k + mx]

• func longestPalindrome(s string) string {
n := len(s)
f := make([][]bool, n)
for i := range f {
f[i] = make([]bool, n)
for j := range f[i] {
f[i][j] = true
}
}
k, mx := 0, 1
for i := n - 2; i >= 0; i-- {
for j := i + 1; j < n; j++ {
f[i][j] = false
if s[i] == s[j] {
f[i][j] = f[i+1][j-1]
if f[i][j] && mx < j-i+1 {
mx = j - i + 1
k = i
}
}
}
}
return s[k : k+mx]
}

• function longestPalindrome(s: string): string {
const n = s.length;
const f: boolean[][] = Array(n)
.fill(0)
.map(() => Array(n).fill(true));
let k = 0;
let mx = 1;
for (let i = n - 2; i >= 0; --i) {
for (let j = i + 1; j < n; ++j) {
f[i][j] = false;
if (s[i] === s[j]) {
f[i][j] = f[i + 1][j - 1];
if (f[i][j] && mx < j - i + 1) {
mx = j - i + 1;
k = i;
}
}
}
}
return s.slice(k, k + mx);
}

• /**
* @param {string} s
* @return {string}
*/
var longestPalindrome = function (s) {
const n = s.length;
const f = Array(n)
.fill(0)
.map(() => Array(n).fill(true));
let k = 0;
let mx = 1;
for (let i = n - 2; i >= 0; --i) {
for (let j = i + 1; j < n; ++j) {
f[i][j] = false;
if (s[i] === s[j]) {
f[i][j] = f[i + 1][j - 1];
if (f[i][j] && mx < j - i + 1) {
mx = j - i + 1;
k = i;
}
}
}
}
return s.slice(k, k + mx);
};

• public class Solution {
public string LongestPalindrome(string s) {
int n = s.Length;
bool[,] f = new bool[n, n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; ++j) {
f[i, j] = true;
}
}
int k = 0, mx = 1;
for (int i = n - 2; i >= 0; --i) {
for (int j = i + 1; j < n; ++j) {
f[i, j] = false;
if (s[i] == s[j]) {
f[i, j] = f[i + 1, j - 1];
if (f[i, j] && mx < j - i + 1) {
mx = j - i + 1;
k = i;
}
}
}
}
return s.Substring(k, mx);
}
}

• import std/sequtils

proc longestPalindrome(s: string): string =
let n: int = s.len()
var
dp = newSeqWith[bool](n, newSeqWith[bool](n, false))
start: int = 0
mx: int = 1

for j in 0 ..< n:
for i in 0 .. j:
if j - i < 2:
dp[i][j] = s[i] == s[j]
else:
dp[i][j] = dp[i + 1][j - 1] and s[i] == s[j]

if dp[i][j] and mx < j - i + 1:
start = i
mx = j - i + 1

result = s[start ..< start+mx]

# Driver Code

• impl Solution {
pub fn longest_palindrome(s: String) -> String {
let (n, mut ans) = (s.len(), &s[..1]);
let mut dp = vec![vec![false; n]; n];
let data: Vec<char> = s.chars().collect();

for end in 1..n {
for start in 0..=end {
if data[start] == data[end] {
dp[start][end] = end - start < 2 || dp[start + 1][end - 1];
if dp[start][end] && end - start + 1 > ans.len() {
ans = &s[start..=end];
}
}
}
}
ans.to_string()
}
}

• class Solution {
/**
* @param string $s * @return string */ function longestPalindrome($s) {
$start = 0;$maxLength = 0;

for ($i = 0;$i < strlen($s);$i++) {
$len1 =$this->expandFromCenter($s,$i, $i);$len2 = $this->expandFromCenter($s, $i,$i + 1);

$len = max($len1, $len2); if ($len > $maxLength) {$start = $i - intval(($len - 1) / 2);
$maxLength =$len;
}
}

return substr($s,$start, $maxLength); } function expandFromCenter($s, $left,$right) {
while ($left >= 0 &&$right < strlen($s) &&$s[$left] ===$s[$right]) {$left--;
$right++; } return$right - \$left - 1;
}
}