# 1745.palindrome-partitioning-iv

## Statement

• Difficulty: Hard
• Tag: `字符串` `动态规划`

``````输入：s = "abcbdd"

``````

``````输入：s = "bcbddxy"

``````

• `3 <= s.length <= 2000`
• `s`​​​​​​ 只包含小写英文字母。

Given a string `s`, return `true` if it is possible to split the string `s` into three non-empty palindromic substrings. Otherwise, return `false`.​​​​​

A string is said to be palindrome if it the same string when reversed.

Example 1:

``````Input: s = "abcbdd"
Output: true
Explanation: "abcbdd" = "a" + "bcb" + "dd", and all three substrings are palindromes.
``````

Example 2:

``````Input: s = "bcbddxy"
Output: false
Explanation: s cannot be split into 3 palindromes.
``````

Constraints:

• `3 <= s.length <= 2000`
• `s`​​​​​​ consists only of 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

const int N = 1e5 + 10;

struct Manacher {
int len, l;
char Ma[N << 1];
int Mp[N << 1];
void work(const char *s) {
len = strlen(s);
l = 0;
Ma[l++] = '\$';
Ma[l++] = '#';
for (int i = 0; i < len; ++i) {
Ma[l++] = s[i];
Ma[l++] = '#';
}
Ma[l] = 0;
int mx = 0, id = 0;
for (int i = 0; i < l; ++i) {
Mp[i] = mx > i ? min(Mp[2 * id - i], mx - i) : 1;
while ((i - Mp[i] >= 0) && Ma[i + Mp[i]] == Ma[i - Mp[i]]) Mp[i]++;
if (i + Mp[i] > mx) {
mx = i + Mp[i];
id = i;
}
}
}
bool check(int l, int r) {
int il = (l + 1) * 2, ir = (r + 1) * 2;
int mid = (il + ir) / 2;
int len = (r - l + 2) / 2;
return (Mp[mid] / 2) >= len;
}
} man;

class Solution {
public:
bool checkPartitioning(string s) {
int n = s.size();
man.work(s.c_str());

for (int i = 0; i < n; i++) {
for (int j = n - 1; j > i + 1; j--) {
if (man.check(0, i) && man.check(i + 1, j - 1) && man.check(j, n - 1)) {
return true;
}
}
}

return false;
}
};

#ifdef LOCAL

int main() {
return 0;
}

#endif
``````