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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.
1def minCut(s: str) -> int:
2 n = len(s)
3 palindrome = [[False] * n for _ in range(n)]
4 cuts = [0] * n
5
6 for i in range(n):
7 min_cut = i
8 for j in range(i + 1):
9 if s[j] == s[i] and (i - j <= 2 or palindrome[j + 1][i - 1]):
10 palindrome[j][i] = True
11 min_cut = 0 if j == 0 else min(min_cut, cuts[j - 1] + 1)
12
13 cuts[i] = min_cut
14
15 return cuts[n - 1]
16
17# Example usage:
18s = "aab"
19print(f"Minimum cuts needed for Palindrome Partitioning: {minCut(s)}")
20In Python, the implementation leverages list comprehension to initialize the palindrome table. The nested loops iterate through the string and update the palindrome conditions and cuts in a similar manner to other languages.
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 <vector>
#include <string>
#include <climits>
using namespace std;
void extendPalindromeAndCalculateCuts(const string &s, int start, int end, vector<int> &dp) {
while (start >= 0 && end < s.size() && s[start] == s[end]) {
dp[end] = min(dp[end], (start == 0 ? 0 : dp[start - 1] + 1));
start--;
end++;
}
}
int minCut(string s) {
int n = s.size();
vector<int> dp(n);
for (int i = 0; i < n; i++) dp[i] = i;
for (int center = 0; center < n; center++) {
extendPalindromeAndCalculateCuts(s, center, center, dp);
extendPalindromeAndCalculateCuts(s, center, center + 1, dp);
}
return dp[n-1];
}
int main() {
string s = "aab";
cout << "Minimum cuts using center expansion: " << minCut(s) << endl;
return 0;
}In C++, this solution optimizes by expanding around each character and middle pair of characters, determining how to best minimize cuts required leveraging the palindrome nature of these centers.