132. Palindrome Partitioning II

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 \(s\).

Python3JavaC++GoTypeScriptC#

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]

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];
    }
}

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];
    }
};

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]
}

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];
}

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];
    }
}

132. Palindrome Partitioning II

Description

Solutions

Solution 1: Dynamic Programming

Comments