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This approach involves using two DP tables. One to check if a substring is a palindrome and another to compute the minimum cuts required.
We maintain a 2D boolean table where palindrome[i][j] is true if the substring s[i...j] is a palindrome. Using this table, we calculate the minimum number of palindrome partitions.
Time Complexity: O(n^2) due to filling the palindrome table and computing cuts.
Space Complexity: O(n^2) as well for the DP tables storage.
1class Solution {
2 public int minCut(String s) {
3 int n = s.length();
4 boolean[][] palindrome = new boolean[n][n];
5 int[] cuts = new int[n];
6
7 for (int i = 0; i < n; i++) {
8 int minCut = i;
9 for (int j = 0; j <= i; j++) {
10 if (s.charAt(j) == s.charAt(i) && (i - j <= 2 || palindrome[j + 1][i - 1])) {
11 palindrome[j][i] = true;
12 minCut = j == 0 ? 0 : Math.min(minCut, cuts[j - 1] + 1);
13 }
14 }
15 cuts[i] = minCut;
16 }
17
18 return cuts[n - 1];
19 }
20
21 public static void main(String[] args) {
22 Solution sol = new Solution();
23 String s = "aab";
24 System.out.println("Minimum cuts needed for Palindrome Partitioning: " + sol.minCut(s));
25 }
26}
27The Java solution is structurally similar to the C++ version, where a 2D array is used to keep track of whether substrings form palindromes. The cuts array is populated based on the palindrome checks ensuring minimum cuts are calculated effectively.
This approach utilizes the idea of expanding around potential palindrome centers, combined with a memoization strategy to store minimum cuts. It significantly reduces redundant calculations by only considering centers and keeping track of the best solutions observed so far.
Time Complexity: O(n^2) due to potentially expanding and updating cuts for each center.
Space Complexity: O(n) focused on the cuts array.
1
JavaScript implementation leverages functions to manage palindrome center expansion effectively for minimizing cuts, leveraging indexed tracking that is optimized per aligned symmetry checks.