README_EN.md

132. Palindrome Partitioning II

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Description

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.

Solutions

Python3

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()
        return ans

class Solution:
    def minCut(self, s: str) -> int:
        n = len(s)
        dp1 = [[False] * n for _ in range(n)]
        for i in range(n - 1, -1, -1):
            for j in range(i, n):
                dp1[i][j] = s[i] == s[j] and (j - i < 3 or dp1[i + 1][j - 1])
        dp2 = [0] * n
        for i in range(n):
            if not dp1[0][i]:
                dp2[i] = i
                for j in range(1, i + 1):
                    if dp1[j][i]:
                        dp2[i] = min(dp2[i], dp2[j - 1] + 1)
        return dp2[-1]

Java

class Solution {
    private boolean[][] g;
    private int[] f;
    private String s;
    private int n;

    public int minCut(String s) {
        n = s.length();
        g = new boolean[n][n];
        for (var e : g) {
            Arrays.fill(e, true);
        }
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                g[i][j] = s.charAt(i) == s.charAt(j) && g[i + 1][j - 1];
            }
        }
        this.s = s;
        f = new int[n];
        Arrays.fill(f, -1);
        return dfs(0);
    }

    private int dfs(int i) {
        if (i >= n - 1) {
            return 0;
        }
        if (f[i] != -1) {
            return f[i];
        }
        int ans = Integer.MAX_VALUE;
        for (int j = i; j < n; ++j) {
            if (g[i][j]) {
                ans = Math.min(ans, dfs(j + 1) + (j < n - 1 ? 1 : 0));
            }
        }
        f[i] = ans;
        return ans;
    }
}

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];
    }
}

Go

func minCut(s string) int {
	n := len(s)
	f := make([]int, n)
	g := make([][]bool, n)
	for i := range g {
		f[i] = -1
		g[i] = make([]bool, n)
		for j := range g[i] {
			g[i][j] = true
		}
	}
	for i := n - 1; i >= 0; i-- {
		for j := i + 1; j < n; j++ {
			g[i][j] = s[i] == s[j] && g[i+1][j-1]
		}
	}
	var dfs func(i int) int
	dfs = func(i int) int {
		if i >= n-1 {
			return 0
		}
		if f[i] != -1 {
			return f[i]
		}
		ans := math.MaxInt32
		for j := i; j < n; j++ {
			if g[i][j] {
				t := 1
				if j == n-1 {
					t = 0
				}
				ans = min(ans, dfs(j+1)+t)
			}
		}
		f[i] = ans
		return ans
	}
	return dfs(0)
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

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
}

C++

class Solution {
public:
    int minCut(string s) {
        int n = s.size();
        vector<vector<bool>> g(n, vector<bool>(n, true));
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                g[i][j] = s[i] == s[j] && g[i + 1][j - 1];
            }
        }
        vector<int> f(n, -1);
        function<int(int)> dfs;
        dfs = [&](int i) {
            if (i >= n - 1) return 0;
            if (f[i] != -1) return f[i];
            int ans = INT_MAX;
            for (int j = i; j < n; ++j) {
                if (g[i][j]) {
                    ans = min(ans, dfs(j + 1) + (j < n - 1));
                }
            }
            f[i] = ans;
            return ans;
        };
        return dfs(0);
    }
};

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];
    }
};

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