132. Palindrome Partitioning II
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.
class Solution {
public int minCut(String s) {
char[] c = s.toCharArray();
int n = c.length;
int[] cut = new int[n];
boolean[][] pal = new boolean[n][n];
for(int i = 0; i < n; i++) {
int min = i;
for(int j = 0; j <= i; j++) {
if(c[j] == c[i] && (j + 1 > i - 1 || pal[j + 1][i - 1])) {
pal[j][i] = true;
min = j == 0 ? 0 : Math.min(min, cut[j - 1] + 1);
}
}
cut[i] = min;
}
return cut[n - 1];
}
}
This can be solved by two points:
- cut[i] is the minimum of cut[j - 1] + 1 (j <= i), if [j, i] is palindrome.
- If [j, i] is palindrome, [j + 1, i - 1] is palindrome, and c[j] == c[i].
The 2nd point reminds us of using dp (caching).
a b a | c c
j i
j-1 | [j, i] is palindrome
cut(j-1) + 1