131. Palindrome Partitioning
Description
Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s.
Example 1:
Input: s = "aab" Output: [["a","a","b"],["aa","b"]]
Example 2:
Input: s = "a" Output: [["a"]]
Constraints:
1 <= s.length <= 16scontains only lowercase English letters.
Solutions
Solution 1: Preprocessing + DFS (Backtracking)
We can use dynamic programming to preprocess whether any substring in the string is a palindrome, i.e., $f[i][j]$ indicates whether the substring $s[i..j]$ is a palindrome.
Next, we design a function $dfs(i)$, which represents starting from the $i$-th character of the string and partitioning it into several palindromic substrings, with the current partition scheme being $t$.
If $i = |s|$, it means the partitioning is complete, and we add $t$ to the answer array and then return.
Otherwise, we can start from $i$ and enumerate the end position $j$ from small to large. If $s[i..j]$ is a palindrome, we add $s[i..j]$ to $t$, then continue to recursively call $dfs(j+1)$. When backtracking, we need to pop $s[i..j]$.
The time complexity is $O(n \times 2^n)$, and the space complexity is $O(n^2)$. Here, $n$ is the length of the string.
Python3
class Solution:
def partition(self, s: str) -> List[List[str]]:
def dfs(i: int):
if i == n:
ans.append(t[:])
return
for j in range(i, n):
if f[i][j]:
t.append(s[i : j + 1])
dfs(j + 1)
t.pop()
n = len(s)
f = [[True] * n for _ in range(n)]
for i in range(n - 1, -1, -1):
for j in range(i + 1, n):
f[i][j] = s[i] == s[j] and f[i + 1][j - 1]
ans = []
t = []
dfs(0)
return ans
Java
class Solution {
private int n;
private String s;
private boolean[][] f;
private List<String> t = new ArrayList<>();
private List<List<String>> ans = new ArrayList<>();
public List<List<String>> partition(String s) {
n = s.length();
f = new boolean[n][n];
for (int i = 0; i < n; ++i) {
Arrays.fill(f[i], true);
}
for (int i = n - 1; i >= 0; --i) {
for (int j = i + 1; j < n; ++j) {
f[i][j] = s.charAt(i) == s.charAt(j) && f[i + 1][j - 1];
}
}
this.s = s;
dfs(0);
return ans;
}
private void dfs(int i) {
if (i == s.length()) {
ans.add(new ArrayList<>(t));
return;
}
for (int j = i; j < n; ++j) {
if (f[i][j]) {
t.add(s.substring(i, j + 1));
dfs(j + 1);
t.remove(t.size() - 1);
}
}
}
}
C++
class Solution {
public:
vector<vector<string>> partition(string s) {
int n = s.size();
bool f[n][n];
memset(f, true, sizeof(f));
for (int i = n - 1; i >= 0; --i) {
for (int j = i + 1; j < n; ++j) {
f[i][j] = s[i] == s[j] && f[i + 1][j - 1];
}
}
vector<vector<string>> ans;
vector<string> t;
auto dfs = [&](this auto&& dfs, int i) -> void {
if (i == n) {
ans.emplace_back(t);
return;
}
for (int j = i; j < n; ++j) {
if (f[i][j]) {
t.emplace_back(s.substr(i, j - i + 1));
dfs(j + 1);
t.pop_back();
}
}
};
dfs(0);
return ans;
}
};
Go
func partition(s string) (ans [][]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
}
}
for i := n - 1; i >= 0; i-- {
for j := i + 1; j < n; j++ {
f[i][j] = s[i] == s[j] && f[i+1][j-1]
}
}
t := []string{}
var dfs func(int)
dfs = func(i int) {
if i == n {
ans = append(ans, append([]string(nil), t...))
return
}
for j := i; j < n; j++ {
if f[i][j] {
t = append(t, s[i:j+1])
dfs(j + 1)
t = t[:len(t)-1]
}
}
}
dfs(0)
return
}
TypeScript
function partition(s: string): string[][] {
const n = s.length;
const f: boolean[][] = Array.from({ length: n }, () => Array(n).fill(true));
for (let i = n - 1; i >= 0; --i) {
for (let j = i + 1; j < n; ++j) {
f[i][j] = s[i] === s[j] && f[i + 1][j - 1];
}
}
const ans: string[][] = [];
const t: string[] = [];
const dfs = (i: number) => {
if (i === n) {
ans.push(t.slice());
return;
}
for (let j = i; j < n; ++j) {
if (f[i][j]) {
t.push(s.slice(i, j + 1));
dfs(j + 1);
t.pop();
}
}
};
dfs(0);
return ans;
}
C#
public class Solution {
private int n;
private string s;
private bool[,] f;
private IList<IList<string>> ans = new List<IList<string>>();
private IList<string> t = new List<string>();
public IList<IList<string>> Partition(string s) {
n = s.Length;
this.s = s;
f = new bool[n, n];
for (int i = 0; i < n; ++i) {
for (int j = 0; j <= i; ++j) {
f[i, j] = true;
}
}
for (int i = n - 1; i >= 0; --i) {
for (int j = i + 1; j < n; ++j) {
f[i, j] = s[i] == s[j] && f[i + 1, j - 1];
}
}
dfs(0);
return ans;
}
private void dfs(int i) {
if (i == n) {
ans.Add(new List<string>(t));
return;
}
for (int j = i; j < n; ++j) {
if (f[i, j]) {
t.Add(s.Substring(i, j + 1 - i));
dfs(j + 1);
t.RemoveAt(t.Count - 1);
}
}
}
}