# [leetcode] Palindrome Partitioning II

Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

For example, given s = `"aab"`,
Return `1` since the palindrome partitioning `["aa","b"]` could be produced using 1 cut.

Dynamic programming.

dp[i] stores the minimum cut of s.substr(0, i + 1).

o(n^3) in time, o(n) in space solution

```class Solution {
public:
int minCut(string s) {
int dp[s.size()];
if(s.size() <= 1) return 0;
dp[0] = 0;
for(int i = 1; i < s.size(); i++){
dp[i] = s.size(); //max value
for(int j = i; j > 0; j--){
if(dp[j - 1] + 1 >= dp[i]) continue;
string tmps = s.substr(j, i - j + 1);
if(isPalindrome(tmps)){
dp[i] = dp[j - 1] + 1 < dp[i] ? dp[j - 1] + 1 : dp[i];
}
}
//j == 0
if(isPalindrome(s.substr(0, i + 1))){
dp[i] = 0;
}
}
return dp[s.size() - 1];
}
bool isPalindrome(string s){
//o(n)
//extreme case
if(s.size() == 1) return true;
int i, j;
if(s.size() % 2 == 0){
//even
i = s.size() / 2 - 1;
j = i + 1;
}
else{
//odd
i = s.size() / 2 - 1;
j = i + 2;
}
while(i >= 0){
if(s[i] != s[j]) return false;
i--;
j++;
}
return true;
}
};```

o(n^2) in time, o(n^2) in space solution.

Construct a palindrome matrix, o(n^2),

Find min cut, o(n^2)

```class Solution {
public:
int minCut(string s) {
int dp[s.size()];
bool matrix[s.size()][s.size()]; //is palindrome
if(s.size() <= 1) return 0;
dp[0] = 0;

memset(matrix, false, sizeof(matrix));
for(int i = 0; i < s.size(); i++){
matrix[i][i] = true;
}
//o(n^2)
for(int i = 0; i < s.size(); i++){
for(int j = i + 1; j < s.size(); j++){
if(s[i] == s[j] && (j - i == 1 || matrix[i + 1][j - 1] == true)){
matrix[i][j] = true;
}
}
for(int j = i - 1; j >= 0 ; j--){
if(s[j] == s[i] && (i - j == 1 || matrix[j + 1][i - 1] == true)){
matrix[j][i] = true;
}
}
}

for(int i = 1; i < s.size(); i++){
dp[i] = s.size(); //max value
for(int j = i; j > 0; j--){
if(dp[j - 1] + 1 >= dp[i]) continue;
if(matrix[j][i] == true){
dp[i] = dp[j - 1] + 1 < dp[i] ? dp[j - 1] + 1 : dp[i];
}
}
//j == 0
if(matrix[0][i] == true){
dp[i] = 0;
}
}
return dp[s.size() - 1];
}
};```

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