Sponsored
Sponsored
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.
The 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#include <stdio.h>
2#include <string.h>
3#include <limits.h>
4
5int extendPalindromeAndCalculateCuts(const char *s, int start, int end, int *dp) {
6 while (start >= 0 && end < strlen(s) && s[start] == s[end]) {
7 dp[end] = (start == 0) ? 0 : ((dp[end] < dp[start - 1] + 1) ? dp[end] : dp[start - 1] + 1);
8 start--;
9 end++;
10 }
11 return dp[end-1];
12}
13
14int minCut(char *s) {
15 int n = strlen(s);
16 int dp[n];
17 for (int i = 0; i < n; i++) dp[i] = i;
18 for (int center = 0; center < n; center++) {
19 extendPalindromeAndCalculateCuts(s, center, center, dp);
20 extendPalindromeAndCalculateCuts(s, center, center + 1, dp);
21 }
22 return dp[n-1];
23}
24
25int main() {
26 char s[] = "aab";
27 printf("Minimum cuts using center expansion: %d\n", minCut(s));
28 return 0;
29}In C, the memoization version checks around potential palindrome centers and expands outwards. It maintains an array for the minimum cuts needed, updating it as palindromes are detected with less cutting as the criteria.
Solve with full IDE support and test cases