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.
Test Cases
Example 1:
Input: s = "aab"
Output: 1
Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.
Example 2:
Input: s = "abaa"
Output: 1
Explanation: The palindrome partitioning ["aba","a"] could be produced using 1 cut.
Solution
class Solution {
public:
int minCut(string s) {
int n =s.length();
bool dp[n][n];
memset(dp, 0, sizeof(dp));
int cut[n];
for(int i=0; i<n; i++) {
int mincut = i;
for(int j=0; j<=i; j++) {
if (s[i] == s[j] && (i-j < 2 || dp[j+1][i-1])) {
dp[j][i] = true;
mincut = min(mincut, j==0 ? 0 : cut[j-1]+1);
}
}
cut[i] = mincut;
}
return cut[n-1];
}
};class Solution {
public int minCut(String s) {
int n =s.length();
boolean[][] dp = new boolean[n][n];
int[] cut= new int[n];
for(int i=0; i<n; i++) {
int mincut = i;
for(int j=0; j<=i; j++) {
if (s.charAt(i) == s.charAt(j) && (i-j < 2 || dp[j+1][i-1])) {
dp[j][i] = true;
mincut = Math.min(mincut, j==0 ? 0 : cut[j-1]+1);
}
}
cut[i] = mincut;
}
return cut[n-1];
}
}