#1147

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Longest Chunked Palindrome Decomposition

Solve the "Longest Chunked Palindrome Decomposition" problem by using dynamic programming and string manipulation techniques.

two pointers string dynamic programming greedy rolling hash

Problem Statement

You are given a string text and need to split it into as many substrings as possible such that each substring is a palindrome. The goal is to maximize the number of these palindromic substrings. The task is to return the largest possible value of k, where k represents the number of palindromic chunks that the string can be split into.

For example, in the string "ghiabcdefhelloadamhelloabcdefghi", the string can be split into 7 palindromic substrings: (ghi)(abcdef)(hello)(adam)(hello)(abcdef)(ghi). The challenge is to efficiently find the maximum number of such chunks using state transition dynamic programming.

Examples

Example 1

Input: text = "ghiabcdefhelloadamhelloabcdefghi"

Output: 7

We can split the string on "(ghi)(abcdef)(hello)(adam)(hello)(abcdef)(ghi)".

Example 2

Input: text = "merchant"

Output: 1

We can split the string on "(merchant)".

Example 3

Input: text = "antaprezatepzapreanta"

Output: 11

We can split the string on "(a)(nt)(a)(pre)(za)(tep)(za)(pre)(a)(nt)(a)".

Constraints

1 <= text.length <= 1000
text consists only of lowercase English characters.

Solution Approach

State Transition Dynamic Programming

Use dynamic programming to break down the problem by progressively checking substrings for potential palindromes. By maintaining an array that tracks the longest palindrome at each point, we can optimize the solution and ensure the substrings are palindromes.

Rolling Hash for Substring Comparison

Implement a rolling hash technique to efficiently compare substrings. This method enables quick verification of whether two substrings are identical, improving the time complexity of checking potential palindrome substrings.

Greedy Approach with Two Pointers

Use the two-pointer technique to efficiently traverse the string, ensuring that substrings are palindromic. This approach, combined with dynamic programming, will maximize the number of palindromic chunks.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the specific approach used. For dynamic programming, it can be O(n^2) where n is the length of the string, while using rolling hash can reduce comparisons to constant time, improving efficiency. Space complexity is also O(n) due to the storage needed for the dynamic programming array and hash values.

What Interviewers Usually Probe

The candidate demonstrates the ability to apply dynamic programming principles effectively.
The candidate shows understanding of optimization techniques like rolling hash for substring comparisons.
The candidate can combine multiple strategies, including two-pointer and greedy approaches, to solve the problem efficiently.

Common Pitfalls or Variants

Common pitfalls

Not properly handling edge cases like very small strings or strings with no palindromes.
Failing to optimize the palindrome check, leading to inefficiencies in time complexity.
Not properly implementing the state transition dynamic programming approach, resulting in incorrect outputs.

Follow-up variants

The problem can be modified to include constraints like requiring each substring to be of even length.
Allowing only certain types of palindromes (e.g., odd-length ones) introduces further complexity.
Changing the input string to a set of characters or more complex symbols could lead to a variation in the approach.

FAQ

What is the main challenge in solving the "Longest Chunked Palindrome Decomposition" problem?

The main challenge is efficiently identifying palindromic substrings and maximizing the number of chunks without exceeding time or space limits. This requires using dynamic programming and rolling hash for efficient substring comparison.

How does the rolling hash technique help in this problem?

Rolling hash allows for efficient comparison of substrings in constant time, significantly improving the time complexity of palindrome verification and speeding up the overall solution.

Can you explain the dynamic programming approach in the "Longest Chunked Palindrome Decomposition" problem?

Dynamic programming helps by storing intermediate results for palindromic substrings, reducing redundant calculations. This makes it possible to build the solution progressively and efficiently.

What is the optimal time complexity for solving this problem?

The optimal time complexity for solving the problem is O(n^2) due to dynamic programming, though rolling hash can reduce the time complexity of substring comparisons.

How does GhostInterview assist in solving the "Longest Chunked Palindrome Decomposition" problem?

GhostInterview provides tailored solutions, offers hints on dynamic programming and rolling hash, and helps you implement efficient techniques for maximizing palindromic chunk splits.

terminal

Solution

Solution 1: Greedy + Two Pointers

We can start from both ends of the string, looking for the shortest, identical, and non-overlapping prefixes and suffixes:

class Solution:
    def longestDecomposition(self, text: str) -> int:
        ans = 0
        i, j = 0, len(text) - 1
        while i <= j:
            k = 1
            ok = False
            while i + k - 1 < j - k + 1:
                if text[i : i + k] == text[j - k + 1 : j + 1]:
                    ans += 2
                    i += k
                    j -= k
                    ok = True
                    break
                k += 1
            if not ok:
                ans += 1
                break
        return ans

Solution 2: String Hash

**String hash** is to map a string of any length to a non-negative integer, and its collision probability is almost $0$. String hash is used to calculate the hash value of a string and quickly determine whether two strings are equal.

class Solution:
    def longestDecomposition(self, text: str) -> int:
        ans = 0
        i, j = 0, len(text) - 1
        while i <= j:
            k = 1
            ok = False
            while i + k - 1 < j - k + 1:
                if text[i : i + k] == text[j - k + 1 : j + 1]:
                    ans += 2
                    i += k
                    j -= k
                    ok = True
                    break
                k += 1
            if not ok:
                ans += 1
                break
        return ans

Continue Practicing

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