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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
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.