Question

link

Stats

Frequency	3
Difficulty	4
Adjusted Difficulty	5
Time to use	--------

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

Analysis

This is DP, but not traditional DP question

IT IS DOUBLE DP!

It is very hard for me to even understand the solution, but I have found a great analysis from peking2’s blog.

这题一般人一看就是DP，DP公式也很容易推出，算是一道简单的DP。

dp(i) = min( 1+dp(j+1), if substring(i,j) is palindrome)

从以上的分析时间复杂度为O(n³), 主要是因为检查回文也需要O(n)的时间。因此这题有意思的一点就是如何降低时间复杂度到O(n²)？

其实这题是两个DP混杂在了一起，这也是这道题最有意思的地方。另外一个DP就是跟检查回文有关了，公式如下

dp(i)(j)=true if s(i)==s(j) && dp(i+1)(j-1)

也就是说，你要检查一个回文只需要知道头尾的字符相等，并且中间的字串已经成为了回文即可。O(1)复杂度。

A more detailed analysis is available in this blog.

[Thoughts]
凡是求最优解的，一般都是走DP的路线。这一题也不例外。首先求dp函数，

定义函数
D[i,n] = 区间[i,n]之间最小的cut数，n为字符串长度

 a   b   a   b   b   b   a   b   b   a   b   a
                     i                                  n
如果现在求[i,n]之间的最优解？应该是多少？简单看一看，至少有下面一个解


 a   b   a   b   b   b   a   b   b   a   b   a
                     i                   j  j+1    n

此时  D[i,n] = min(D[i, j] + D[j+1,n])  i<=j <n。这是个二维的函数，实际写代码时维护比较麻烦。所以要转换成一维DP。如果每次，从i往右扫描，每找到一个回文就算一次DP的话，就可以转换为
D[i] = 区间[i,n]之间最小的cut数，n为字符串长度， 则,

D[i] = min(1+D[j+1] )    i<=j <n

有个转移函数之后，一个问题出现了，就是如何判断[i,j]是否是回文？每次都从i到j比较一遍？太浪费了，这里也是一个DP问题。
定义函数
P[i][j] = true if [i,j]为回文

那么
P[i][j] = str[i] == str[j] && P[i+1][j-1];

Solution

The coding is not easy, especially when 2 DP are written in 1 for-loop.

I wrote many times until I finally achieved the nice and concise solution that you see below.

Code

Doing everything in 1 loop, not an intuitive code.

public int minCut(String s) {
    int len = s.length();
    if (len <= 1) return 0;
    boolean[][] pl = new boolean[len][len];
    int[] dp = new int[len];
    for (int i = len-1; i >= 0; i --) {
        dp[i] = Integer.MAX_VALUE;
        for (int j = i; j < len; j ++) {
            // first set pl[][], then update dp[i]
            if (j - i <= 1) pl[i][j] = s.charAt(i) == s.charAt(j);
            else pl[i][j] = s.charAt(i) == s.charAt(j) & pl[i+1][j-1];
            if (pl[i][j]) {
                if (j == len-1) dp[i] = 0;
                else 
                    dp[i] = Math.min(dp[i], dp[j+1] + 1);
            }
        }
    }
    return dp[0];
}

Updated on July 18th, 2014, written by me.

boolean[][] map = null;

public int minCut(String s) {
    if (s == null || s.length() == 0) {
        return 0;
    }
    map = getMap(s);
    int len = s.length();
    int[] dp = new int[len + 1];
    dp[0] = -1;
    for (int i = 0; i < len; i++) {
        dp[i+1] = Integer.MAX_VALUE;
        for (int j = 0; j <= i; j++) {
            if (map[j][i]) {
                dp[i+1] = Math.min(dp[i+1], dp[j] + 1);
            }
        }
    }
    return dp[len];
}

private boolean[][] getMap(String s) {
    int len = s.length();
    boolean[][] map = new boolean[len][len];
    for (int i = len - 1; i >= 0; i--) {
        for (int j = 0; j < len; j++) {
            if (i > j) {
                continue;
            } else if (i == j) {
                map[i][j] = true;
            } else {
                if (i + 1 == j) {
                    map[i][j] = s.charAt(i) == s.charAt(j);
                } else {
                    map[i][j] = s.charAt(i) == s.charAt(j) && map[i+1][j-1];
                }
            }
        }
    }
    return map;
}