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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.
1#include <stdbool.h>
2#include <stdio.h>
3#include <string.h>
4
5int minCut(char *s) {
6 int n = strlen(s);
7 bool palindrome[n][n];
8 int cuts[n];
9
10 for (int i = 0; i < n; i++) {
11 cuts[i] = i;
12 for (int j = 0; j <= i; j++) {
13 if (s[j] == s[i] && (i - j <= 2 || palindrome[j + 1][i - 1])) {
14 palindrome[j][i] = true;
15 cuts[i] = j == 0 ? 0 : (cuts[j - 1] + 1);
16 }
17 }
18 }
19
20 return cuts[n - 1];
21}
22
23int main() {
24 char s[] = "aab";
25 printf("Minimum cuts needed for Palindrome Partitioning: %d\n", minCut(s));
26 return 0;
27}
28The solution builds a palindrome table by iterating through the string. It checks every substring and stores whether it is a palindrome or not. Then we use this table to compute the minimum cuts for each position in the string. Cuts are initialized assuming each character needs a cut, and the DP table is updated accordingly for each palindrome found.
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.