Palindrome Partitioning II
Logical Breakdown
- [ ] Subproblem Identification:
- [ ] Optimal Substructure:
- [ ] Constraint Handling: (e.g., Modulo \(10^9 + 7\))
- [ ] Optimization: (Matrix Exponentiation if 'A' is huge)
Mathematical Rigor
State Definition
Let \(dp[i]\) be the state for the \(i\)-th subproblem.
Recurrence Relation
Visualization
Complexity Analysis
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Iterative DP | \(O(N)\) | \(O(1)\) |
Code Reference
package com.dsa.dynamic_programming;
import java.util.*;
/* * ============================================================================================ * PROBLEM: Palindrome Partitioning II * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Given a string s, partition s such that every substring of the partition is a palindrome. * Return the MINIMUM cuts needed. * * --------------------- * 2. Logical Breakdown * --------------------- * This requires two layers of logic: * * Part A: Fast Palindrome Check * Precompute a boolean table isPal[i][j] which is true if s[i...j] is a palindrome. * This takes O(N^2) time. * * Part B: Min Cuts DP * Let dp[i] be the min cuts needed for the prefix s[0...i]. * * Recurrence: * To calculate dp[i]: * Iterate j from 0 to i. * If s[j...i] is a palindrome (check isPal[j][i]): * Then we can make a cut at j-1. * Possible Cuts = (j == 0 ? 0 : dp[j-1] + 1) * dp[i] = min(dp[i], Possible Cuts) * * --------------------- * 3. Complexity * --------------------- * Time Complexity: O(N^2) * Space Complexity: O(N^2) * * ============================================================================================ / public class PalindromePartitioningII { public int solve(int A) { // TODO: Implement solution return 0; }
public int minCut(String ab) {
return 1;
}
}