Question

Formatted question description: https://leetcode.ca/all/132.html

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.

Example 1:

Input: s = "aab"
Output: 1
Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.

Example 2:

Input: s = "a"
Output: 0

Example 3:

Input: s = "ab"
Output: 1

Constraints:

1 <= s.length <= 2000
s consists of lowercase English letters only.

Algorithm

One-dimensional dp array, where dp[i] represents the minimum number of divisions in the range of substring [0, i].

Use a two-dimensional dp array p, where p[i][j] indicates whether the substring in the interval [i, j] is a palindrome, and the state transition equation is p[i][j] = (s[i] == s[j]) && p[i+1][j-1], where p[i][j] = true if [i, j] is a palindrome.

Code

class Solution {
    public int minCut(String s) {
        int n = s.length();
        boolean[][] dp1 = new boolean[n][n];
        for (int i = n - 1; i >= 0; i--) {
            for (int j = i; j < n; j++) {
                dp1[i][j] = s.charAt(i) == s.charAt(j) && (j - i < 3 || dp1[i + 1][j - 1]);
            }
        }
        int[] dp2 = new int[n];
        for (int i = 0; i < n; i++) {
            if (!dp1[0][i]) {
                dp2[i] = i;
                for (int j = 1; j <= i; j++) {
                    if (dp1[j][i]) {
                        dp2[i] = Math.min(dp2[i], dp2[j - 1] + 1);
                    }
                }
            }
        }
        return dp2[n - 1];
    }
}

############

public class Palindrome_Partitioning_II {

    public class Solution {
        public int minCut(String s) {

            if (s == null || s.length() == 0) {
                return -1;
            }

            // 'abcba', when reaching 'c', minCut=2
            //          when reaching 2nd 'b', minCut=1
            //          when reaching 2nd 'a', minCut=0
            // so, if a substring is palindrome, cut is prev_segment+1

            // set up a fast lookup for if palindrome from i to j
            boolean[][] isPal = new boolean[s.length()][s.length()];
            for (int i = 0; i < s.length(); i++) {
                for (int j = i; j >= 0; j--) {
                    if (i == j || (s.charAt(i) == s.charAt(j) && (i - j == 1 || isPal[j + 1][i - 1]))) {
                        isPal[j][i] = true; // j is first, then i
                    }
                }
            }

            // set up min count array, worst case, each char is a palindrome
            int[] minCut = new int[s.length()];
            for (int i = 0; i < s.length(); i++) {
                minCut[i] = i;
            }

            // how to cound, eg: 'aabcba', 'aabcbaz'
            // @note: if a substring is palindrome, cut is prev_segment+1
            for (int i = 0; i < s.length(); i++) {
                for (int j = 0; j < s.length(); j++) {
                    if (isPal[i][j]) {
                        if (i - 1 >= 0) { // @note: eg, 'aabcba'
                            minCut[j] = Math.min(minCut[j], 1 + minCut[i - 1]);
                        } else { // i.e., i==0. eg, 'abcba'
                            minCut[j] = 0;
                        }
                    }
                }
            }

            int min = minCut[s.length() - 1];
            return min;
        }
    }
}

class Solution {
public:
    int minCut(string s) {
        int n = s.size();
        vector<vector<bool>> dp1(n, vector<bool>(n));
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i; j < n; ++j) {
                dp1[i][j] = s[i] == s[j] && (j - i < 3 || dp1[i + 1][j - 1]);
            }
        }
        vector<int> dp2(n);
        for (int i = 0; i < n; ++i) {
            if (!dp1[0][i]) {
                dp2[i] = i;
                for (int j = 1; j <= i; ++j) {
                    if (dp1[j][i]) {
                        dp2[i] = min(dp2[i], dp2[j - 1] + 1);
                    }
                }
            }
        }
        return dp2[n - 1];
    }
};

class Solution:
    def minCut(self, s: str) -> int:
        pal = [[False for j in range(0, len(s))] for i in range(0, len(s))]
        dp = [len(s) for _ in range(0, len(s) + 1)]
        for i in range(0, len(s)):
            for j in range(0, i + 1):
                # also ok if, (j + 1 >= i - 1)
                if (s[i] == s[j]) and ((j + 1 > i - 1) or (pal[i - 1][j + 1])):
                    pal[i][j] = True
                    dp[i + 1] = min(dp[i + 1], dp[j] + 1) if j != 0 else 0
        return dp[-1]


class Solution:
    def minCut(self, s: str) -> int:
        @cache
        def dfs(i):
            if i >= n - 1:
                return 0
            ans = inf
            for j in range(i, n):
                if g[i][j]:
                    ans = min(ans, dfs(j + 1) + (j < n - 1))
            return ans

        n = len(s)
        g = [[True] * n for _ in range(n)]
        for i in range(n - 1, -1, -1):
            for j in range(i + 1, n):
                g[i][j] = s[i] == s[j] and g[i + 1][j - 1]
        ans = dfs(0)
        dfs.cache_clear() # nice, clear cache !!!
        return ans


class Solution:
    def minCut(self, s: str) -> int:
        n = len(s)
        is_pal = [[False] * n for _ in range(n)]
        for i in range(n - 1, -1, -1):
            for j in range(i, n):
                is_pal[i][j] = s[i] == s[j] and (j - i < 3 or is_pal[i + 1][j - 1])
        min_cut = [i for i in range(n)]
        for i in range(n):
            for j in range(0, i + 1):
                if is_pal[j][i]:
                    min_cut[i] = min(min_cut[i], min_cut[j - 1] + 1) if j > 0 else 0
        return min_cut[-1]

func minCut(s string) int {
	n := len(s)
	dp1 := make([][]bool, n)
	for i := 0; i < n; i++ {
		dp1[i] = make([]bool, n)
	}
	for i := n - 1; i >= 0; i-- {
		for j := i; j < n; j++ {
			dp1[i][j] = s[i] == s[j] && (j-i < 3 || dp1[i+1][j-1])
		}
	}
	dp2 := make([]int, n)
	for i := 0; i < n; i++ {
		if !dp1[0][i] {
			dp2[i] = i
			for j := 1; j <= i; j++ {
				if dp1[j][i] {
					dp2[i] = min(dp2[i], dp2[j-1]+1)
				}
			}
		}
	}
	return dp2[n-1]
}

func min(x, y int) int {
	if x < y {
		return x
	}
	return y
}

using System;
using System.Collections.Generic;

public class Solution {
    public int MinCut(string s) {
        if (s.Length == 0) return 0;
            
        var paths = new List<int>[s.Length];
        for (var i = 0; i < s.Length; ++i)
        {
            int j, k;
            for (j = i, k = i + 1; j >= 0 && k < s.Length; --j, ++k)
            {
                if (s[j] == s[k])
                {
                    if (paths[k] == null)
                    {
                        paths[k] = new List<int>();
                    }
                    paths[k].Add(j - 1);
                }
                else
                {
                    break;
                }
            }
            for (j = i, k = i; j >= 0 && k < s.Length; --j, ++k)
            {
                if (s[j] == s[k])
                {
                    if (paths[k] == null)
                    {
                        paths[k] = new List<int>();
                    }
                    paths[k].Add(j - 1);
                }
                else
                {
                    break;
                }
            }
        }
        
        var partCount = new int[s.Length];
        for (var i = 0; i < s.Length; ++i)
        {
            partCount[i] = int.MaxValue;
            if (paths[i] != null)
            {
                foreach (var path in paths[i])
                {
                    if (path < 0)
                    {
                        partCount[i] = 0;
                        break;
                    }
                    else
                    {
                        partCount[i] = Math.Min(partCount[i], partCount[path]);
                    }
                }
            }
            ++partCount[i];
        }
        return partCount[s.Length - 1] - 1;        
    }
}

132 - Palindrome Partitioning II

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