README_EN.md

comments	difficulty	edit_url	tags
true	Hard	https://github.com/doocs/leetcode/edit/main/solution/0100-0199/0132.Palindrome%20Partitioning%20II/README_EN.md	String Dynamic Programming

132. Palindrome Partitioning II

中文文档

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

Solution 1: Dynamic Programming

First, we preprocess the string $s$ to determine whether each substring $s[i..j]$ is a palindrome, and record this in a 2D array $g[i][j]$, where $g[i][j]$ indicates whether the substring $s[i..j]$ is a palindrome.

Next, we define $f[i]$ to represent the minimum number of cuts needed for the substring $s[0..i-1]$. Initially, $f[i] = i$.

Next, we consider how to transition the state for $f[i]$. We can enumerate the previous cut point $j$. If the substring $s[j..i]$ is a palindrome, then $f[i]$ can be transitioned from $f[j]$. If $j = 0$, it means that $s[0..i]$ itself is a palindrome, and no cuts are needed, i.e., $f[i] = 0$. Therefore, the state transition equation is as follows:

$$ f[i] = \min_{0 \leq j \leq i} \begin{cases} f[j-1] + 1, & \textit{if}\ g[j][i] = \textit{True} \ 0, & \textit{if}\ g[0][i] = \textit{True} \end{cases} $$

The answer is $f[n]$, where $n$ is the length of the string $s$.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the length of the string $s$.

Python3

class Solution:
    def minCut(self, s: str) -> int:
        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]
        f = list(range(n))
        for i in range(1, n):
            for j in range(i + 1):
                if g[j][i]:
                    f[i] = min(f[i], 1 + f[j - 1] if j else 0)
        return f[-1]

Java

class Solution {
    public int minCut(String s) {
        int n = s.length();
        boolean[][] g = new boolean[n][n];
        for (var row : g) {
            Arrays.fill(row, 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];
            }
        }
        int[] f = new int[n];
        for (int i = 0; i < n; ++i) {
            f[i] = i;
        }
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j <= i; ++j) {
                if (g[j][i]) {
                    f[i] = Math.min(f[i], j > 0 ? 1 + f[j - 1] : 0);
                }
            }
        }
        return f[n - 1];
    }
}

C++

class Solution {
public:
    int minCut(string s) {
        int n = s.size();
        bool g[n][n];
        memset(g, true, sizeof(g));
        for (int i = n - 1; ~i; --i) {
            for (int j = i + 1; j < n; ++j) {
                g[i][j] = s[i] == s[j] && g[i + 1][j - 1];
            }
        }
        int f[n];
        iota(f, f + n, 0);
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j <= i; ++j) {
                if (g[j][i]) {
                    f[i] = min(f[i], j ? 1 + f[j - 1] : 0);
                }
            }
        }
        return f[n - 1];
    }
};

Go

func minCut(s string) int {
	n := len(s)
	g := make([][]bool, n)
	f := make([]int, n)
	for i := range g {
		g[i] = make([]bool, n)
		f[i] = i
		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]
		}
	}
	for i := 1; i < n; i++ {
		for j := 0; j <= i; j++ {
			if g[j][i] {
				if j == 0 {
					f[i] = 0
				} else {
					f[i] = min(f[i], f[j-1]+1)
				}
			}
		}
	}
	return f[n-1]
}

TypeScript

function minCut(s: string): number {
    const n = s.length;
    const g: boolean[][] = Array.from({ length: n }, () => Array(n).fill(true));
    for (let i = n - 1; ~i; --i) {
        for (let j = i + 1; j < n; ++j) {
            g[i][j] = s[i] === s[j] && g[i + 1][j - 1];
        }
    }
    const f: number[] = Array.from({ length: n }, (_, i) => i);
    for (let i = 1; i < n; ++i) {
        for (let j = 0; j <= i; ++j) {
            if (g[j][i]) {
                f[i] = Math.min(f[i], j ? 1 + f[j - 1] : 0);
            }
        }
    }
    return f[n - 1];
}

C#

public class Solution {
    public int MinCut(string s) {
        int n = s.Length;
        bool[,] g = new bool[n,n];
        int[] f = new int[n];
        for (int i = 0; i < n; ++i) {
            f[i] = i;
            for (int j = 0; j < n; ++j) {
                g[i,j] = 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];
            }
        }
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j <= i; ++j) {
                if (g[j,i]) {
                    f[i] = Math.Min(f[i], j > 0 ? 1 + f[j - 1] : 0);
                }
            }
        }
        return f[n - 1];
    }
}