132. 分割回文串 II

给你一个字符串 s，请你将 s 分割成一些子串，使每个子串都是回文。

返回符合要求的 最少分割次数 。

示例 1：

输入：s = "aab"
输出：1
解释：只需一次分割就可将 s 分割成 ["aa","b"] 这样两个回文子串。

示例 2：

输入：s = "a"
输出：0

示例 3：

输入：s = "ab"
输出：1

提示：

• 1 <= s.length <= 2000
• s 仅由小写英文字母组成

use backtrack solution from131. 分割回文串, oops, get tle and sle

class Solution:
    def minCut(self, s: str) -> int:
        res=[]
        def helper(s,tmp):
            if not s:
                res.append(tmp)#空字符说明处理完毕
            for i in range(1,len(s)+1):
                if s[:i]==s[:i][::-1]:#判断回文
                    helper(s[i:],tmp+[s[:i]])#回溯
        
        helper(s,[])
        return min(len(x)-1 for x in res)

class Solution {
    public int minCut(String s) {
        List<List<String>> list = new ArrayList<>();
        backtrack(list, new ArrayList<>(), s, 0);
        int ans=2010;
        for(int i=0;i<list.size();i++){
            ans=Math.min(list.get(i).size(),ans);
        }
        return ans-1;
    }

    public void backtrack(List<List<String>> list, List<String> tempList, String s, int start){
        if(start == s.length())
            list.add(new ArrayList<>(tempList));
        else{
            for(int i = start; i < s.length(); i++){
                if(isPalindrome(s, start, i)){
                    tempList.add(s.substring(start, i + 1));
                    backtrack(list, tempList, s, i + 1);
                    tempList.remove(tempList.size() - 1);
                }
            }
        }
    }

    public boolean isPalindrome(String s, int low, int high){
        while(low < high)
            if(s.charAt(low++) != s.charAt(high--)) return false;
        return true;
    }
}

now we have to come up with a DP solution

The idea is that if c[j+1] to c[i-1] is the number of palindromes and there is c[j]==c[i], then the number of palindromes is extended. Pal[j][i]=true means that c[j] to c[i] are the number of palindromes.

Time complexity O(n^2)

Space complexity O(n^2)

class Solution {
    public int minCut(String s) {
        char[] c=s.toCharArray();
        int n=c.length;
        int[] cut=new int[n];
        boolean[][] pal=new boolean[n][n];
        for(int i=0;i<n;i++){
            int min=i;
            for(int j = 0;j <= i; j++){
                if(c[j]==c[i] && (j+1 > i-1 || pal[j+1][i-1])){
                    pal[j][i]=true;
                    min = j == 0 ? 0 : Math.min(min, cut[j - 1] + 1);
                }
            }
            cut[i] = min;
        }
        return cut[n-1];
    }
}