Palindrome Partitioning II
Problem Description
Question
Given a string s, partition s such that every substring of the partition is a palindrome.
Return the minimum cuts needed for a palindrome partitioning of s.
Example 1:
Input: s = "aab"
Output: 1
Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.
Example 2:
Input: s = "a"
Output: 0
Example 3:
Input: s = "ab"
Output: 1
Constraints:
1 <= s.length <= 2000sconsists of lowercase English letters only.
Solve it on LeetCode
Approaches
1. Dynamic Programming with Palindrome Check
Intuition:
The main idea is to use dynamic programming to keep track of the minimum cuts needed for a substring. We start by calculating if any substring is a palindrome and then use this information to determine the minimum number of cuts needed for each substring.
Approach:
- Use a 2D boolean matrix
palindromewherepalindrome[i][j] = trueif the substrings[i...j]is a palindrome. - Use a 1D array
cutwherecut[i]holds the minimum cuts needed for the substrings[0...i]. - Populate the
palindromematrix using the conditions: s[i] == s[j]andpalindrome[i+1][j-1] == trueif the substring is longer than two.- For each position
i, calculate the minimum cuts by checking if any substrings[j...i]is a palindrome and updatecut[i]accordingly.
Code:
Complexity Analysis
- Time Complexity:
O(n^2), wherenis the length of the string. We are filling up ann x nmatrix. - Space Complexity:
O(n^2)for the palindrome matrix.
2. Optimized Dynamic Programming
Intuition:
We can optimize the solution by reducing unnecessary recalculations and using a more memory-efficient way to determine palindrome substrings.
Approach:
- Instead of using a 2D array to track palindromes, we maintain one array for storing minimum cuts and use an inner loop to decide if a substring is a palindrome on the fly.
- Use two loops to handle both even and odd length center expansions for palindrome checking.
Code:
Complexity Analysis
- Time Complexity:
O(n^2), still iterating potential palindrome centers and expanding. - Space Complexity:
O(n), no extra 2D array. Only using thecutarray for space.