# 131.palindrome-partitioning

## Statement

• Difficulty: Medium
• Tag: `字符串` `动态规划` `回溯`

``````输入：s = "aab"

``````

``````输入：s = "a"

``````

• `1 <= s.length <= 16`
• `s` 仅由小写英文字母组成

• Difficulty: Medium
• Tag: `String` `Dynamic Programming` `Backtracking`

Given a string `s`, partition `s` such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of `s`.

A palindrome string is a string that reads the same backward as forward.

Example 1:

``````Input: s = "aab"
Output: [["a","a","b"],["aa","b"]]
``````

Example 2:

``````Input: s = "a"
Output: [["a"]]
``````

Constraints:

• `1 <= s.length <= 16`
• `s` contains only lowercase English letters.

## Solution

``````#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T& a, const S& b) {
return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T& a, const S& b) {
return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif

class Solution {
public:
bool check(const string& s) {
string t = s;
reverse(all(t));
return s == t;
}

bool checkAll(const vector<string>& ss) {
for (const auto& s : ss) {
if (!check(s)) {
return false;
}
}

return true;
}

void dfs(const string& s, int ix, string t, vector<string> cur, vector<vector<string>>& res) {
if (ix == s.length()) {
if (!t.empty()) {
return;
}

if (checkAll(cur)) {
res.push_back(cur);
}

return;
}

t += s[ix];
dfs(s, ix + 1, t, cur, res);

cur.push_back(t);
dfs(s, ix + 1, "", cur, res);
}

vector<vector<string>> partition(string s) {
vector<vector<string>> res;

dfs(s, 0, "", vector<string>(), res);

return res;
}
};

#ifdef LOCAL

int main() {
return 0;
}

#endif
``````