Hard
Palindrome Partitioning IV — C++
Full explanation · Time O(n^2) · Space O(n)
// Time: O(n^2)
// Space: O(n)
class Solution {
public:
bool checkPartitioning(string s) {
const auto& P = manacher(s);
vector<int> prefix, suffix;
for (int i = 2; i < size(P) - 2; ++i) {
if (i - 1 - P[i] == 0) {
prefix.emplace_back(i);
}
if (i + 1 + P[i] == size(P) - 1) {
suffix.emplace_back(i);
}
}
for (const auto& i : prefix) {
for (const auto& j : suffix) {
int left = i + 1 + P[i], right = j - 1 - P[j];
if (left > right) {
continue;
}
int mid = left + (right - left) / 2;
if (P[mid] >= mid - left) {
return true;
}
}
}
return false;
}
private:
vector<int> manacher(const string& s) {
string T = preProcess(s);
const int n = size(T);
vector<int> P(n);
int C = 0, R = 0;
for (int i = 1; i < n - 1; ++i) {
int i_mirror = 2 * C - i;
P[i] = (R > i) ? min(R - i, P[i_mirror]) : 0;
while (T[i + 1 + P[i]] == T[i - 1 - P[i]]) {
++P[i];
}
if (i + P[i] > R) {
C = i;
R = i + P[i];
}
}
return P;
}
string preProcess(const string& s) {
if (empty(s)) {
return "^$";
}
string ret = "^";
for (int i = 0; i < size(s); ++i) {
ret += "#" + s.substr(i, 1);
}
ret += "#$";
return ret;
}
};
// Time: O(n^2)
// Space: O(n^2)
class Solution2 {
public:
bool checkPartitioning(string s) {
vector<vector<bool>> dp(size(s), vector<bool>(size(s)));
for (int i = size(s) - 1; i >= 0; --i) {
for (int j = i; j < size(s); ++j) {
if (s[i] == s[j] && (j - i < 2 || dp[i + 1][j - 1])) {
dp[i][j] = true;
}
}
}
for (int i = 1; i + 1 < size(s); ++i) {
if (!dp[0][i - 1]) {
continue;
}
for (int j = i + 1; j < size(s); ++j) {
if (!dp[j][size(s) - 1]) {
continue;
}
if (dp[i][j - 1]) {
return true;
}
}
}
return false;
}
};