Palindrome Partitioning II - Video Solutions
DP 53. Palindrome Partitioning - II | Front Partition π₯
Palindrome Partitioning II - Video Solution
Watch 10 video solutions for Palindrome Partitioning II, a hard level problem involving String, Dynamic Programming. This walkthrough by take U forward has 194,138 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
Problem Statement
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.
Approach Overview
Problem Overview: Given a string s, split it into substrings such that every substring is a palindrome. Return the minimum number of cuts required. A single character is already a palindrome, but longer segments must read the same forward and backward.
Approach 1: Dynamic Programming with Palindrome Table (Time: O(n^2), Space: O(n^2))
This method precomputes whether every substring s[i..j] is a palindrome. Use a 2D boolean table where pal[i][j] becomes true if the characters match and the inner substring s[i+1..j-1] is also a palindrome. Once this table is built, compute the minimum cuts using a DP array cuts[i] representing the minimum cuts needed for prefix s[0..i]. For each index, iterate backward to find valid palindrome segments and update cuts[i]. The precomputed palindrome table removes repeated checks, giving an overall O(n^2) time solution. This approach is common in dynamic programming problems involving substring partitions.
Approach 2: Center Expansion with Memoization (Time: O(n^2), Space: O(n))
Instead of storing all palindrome substrings, expand around each center to detect palindromes dynamically. For every index, expand for both odd and even length palindromes using two pointers moving outward. When a palindrome s[l..r] is found, update the minimum cuts needed for position r using previously computed results. A memoized DP array tracks the best cut count for every prefix of the string. This avoids the n Γ n palindrome table while still exploring all valid centers. The technique relies on the symmetry property of palindromes and is a common pattern in string problems involving substring checks.
Recommended for interviews: The dynamic programming approach with a palindrome table is the most expected answer. It clearly separates two ideas: detecting palindromes and minimizing cuts. Interviewers like this solution because the reasoning is easy to explain and guarantees O(n^2) time. The center expansion method is slightly more space efficient and demonstrates deeper understanding of palindrome properties, which can stand out in strong DP interviews.
Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Dynamic Programming with Palindrome Table | O(n^2) | O(n^2) | Standard interview solution when clarity is more important than memory usage |
| Center Expansion with Memoization | O(n^2) | O(n) | When you want to avoid storing the full palindrome table and compute palindromes on the fly |