## Palindrome Partitioning

• Time:O(n \cdot 2^n)
• Space:O(n \cdot 2^n)

## C++

class Solution {
public:
vector<vector<string>> partition(string s) {
vector<vector<string>> ans;
dfs(s, 0, {}, ans);
return ans;
}

private:
void dfs(const string& s, int start, vector<string>&& path,
vector<vector<string>>& ans) {
if (start == s.length()) {
ans.push_back(path);
return;
}

for (int i = start; i < s.length(); ++i)
if (isPalindrome(s, start, i)) {
path.push_back(s.substr(start, i - start + 1));
dfs(s, i + 1, move(path), ans);
path.pop_back();
}
}

bool isPalindrome(const string& s, int l, int r) {
while (l < r)
if (s[l++] != s[r--])
return false;
return true;
}
};


## JAVA

class Solution {
public List<List<String>> partition(String s) {
List<List<String>> ans = new ArrayList<>();
dfs(s, 0, new ArrayList<>(), ans);
return ans;
}

private void dfs(final String s, int start, List<String> path, List<List<String>> ans) {
if (start == s.length()) {
return;
}

for (int i = start; i < s.length(); ++i)
if (isPalindrome(s, start, i)) {
dfs(s, i + 1, path, ans);
path.remove(path.size() - 1);
}
}

private boolean isPalindrome(final String s, int l, int r) {
while (l < r)
if (s.charAt(l++) != s.charAt(r--))
return false;
return true;
}
}


## Python

class Solution:
def partition(self, s: str) -> List[List[str]]:
ans = []

def isPalindrome(s: str) -> bool:
return s == s[::-1]

def dfs(s: str, j: int, path: List[str], ans: List[List[str]]) -> None:
if j == len(s):
ans.append(path)
return

for i in range(j, len(s)):
if isPalindrome(s[j: i + 1]):
dfs(s, i + 1, path + [s[j: i + 1]], ans)

dfs(s, 0, [], ans)
return ans