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 <= 2000s 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.
PythonJavaC++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](code-box)
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];
}
}(code-box)
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];
}
};(code-box)
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]
}(code-box)
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];
}(code-box)
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];
}
}(code-box)
