What is the time complexity of Maximum Number of Non-overlapping Palindrome Substrings?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Maximum Number of Non-overlapping Palindrome Substrings?

This problem uses Two Pointers, String, Dynamic Programming, Greedy. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Number of Non-overlapping Palindrome Substrings?

Insight 1: Dynamic programming is a suitable approach because we can build up the solution incrementally. We can store the maximum number of palindromes we can find up to a certain index in the string. Insight 2: The core idea is to consider each position `i` in the string. At this position, we can either not include any new palindrome substring ending at `i`, or include a palindrome substring of length `k` or `k+1` (if it exists) ending at `i`. We then take the maximum of these possibilities.

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Maximum Number of Non-overlapping Palindrome Substrings - LeetCode 2472 Solution

Q: Which companies ask LeetCode 2472. Maximum Number of Non-overlapping Palindrome Substrings in interviews?

This problem is frequently asked at LinkedIn, SoFi, Walmart Labs.

Maximum Number of Non-overlapping Palindrome Substrings - Complete Solution Guide

Maximum Number of Non-overlapping Palindrome Substrings is LeetCode problem 2472, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a string s and a positive integer k . Select a set of non-overlapping substrings from the string s that satisfy the following conditions: The length of each substring is at least k . Each substring is a palindrome . Return the maximum number of substrings in an optimal selection . A substring is a contiguous sequence of characters within a string. Example 1: Input: s = "abaccdbbd", k = 3 Output: 2 Explanation: We can select the substrings underlined in s = " aba cc dbbd ". Both "ab

Detailed Explanation

The problem asks us to find the maximum number of non-overlapping palindrome substrings within a given string `s`, where each palindrome substring must have a length of at least `k`. We need to select these substrings such that they don't overlap with each other and maximize the total count of selected palindromes.

Solution Approach

The provided solutions use dynamic programming to solve this problem. `dp[i]` represents the maximum number of non-overlapping palindrome substrings ending at index `i-1` of the string `s`. For each index `i`, we consider two cases: either we don't include a new palindrome ending at `i-1` or we do. If we include a palindrome ending at `i-1`, we must consider the possible lengths of such a palindrome, which are `k` and `k+1`. We update `dp[i]` by taking the maximum of `dp[i-1]` (not including a new palindrome), `dp[i-k] + 1` (if a palindrome of length `k` exists ending at `i-1`), and `dp[i-k-1] + 1` (if a palindrome of length `k+1` exists ending at `i-1`). The final answer is stored in `dp[n]`, where `n` is the length of the string.

Step-by-Step Algorithm

Step 1: Initialize a DP array `dp` of size `n+1` with all elements set to 0. `dp[i]` will store the maximum number of non-overlapping palindrome substrings in `s[0...i-1]`.
Step 2: Iterate through the string `s` from index 1 to `n`. For each index `i`, initialize `dp[i]` to `dp[i-1]`. This represents the case where we don't include any new palindrome ending at `i-1`.
Step 3: Check if a palindrome of length `k` exists ending at index `i-1`. Extract the substring `s[i-k:i]` and check if it's a palindrome. If it is, update `dp[i]` to `max(dp[i], dp[i-k] + 1)`.
Step 4: Check if a palindrome of length `k+1` exists ending at index `i-1`. Extract the substring `s[i-k-1:i]` and check if it's a palindrome. If it is, update `dp[i]` to `max(dp[i], dp[i-k-1] + 1)`.
Step 5: After iterating through the entire string, `dp[n]` will contain the maximum number of non-overlapping palindrome substrings. Return `dp[n]`.

Key Insights

Insight 1: Dynamic programming is a suitable approach because we can build up the solution incrementally. We can store the maximum number of palindromes we can find up to a certain index in the string.
Insight 2: The core idea is to consider each position `i` in the string. At this position, we can either not include any new palindrome substring ending at `i`, or include a palindrome substring of length `k` or `k+1` (if it exists) ending at `i`. We then take the maximum of these possibilities.
Insight 3: We only need to check for palindromes of length `k` and `k+1`. Longer palindromes can potentially be broken down into overlapping palindromes of length k and k+1 and not increase our palindrome count, so considering lengths greater than k+1 is redundant.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Two Pointers, String, Dynamic Programming, Greedy.

Companies

Asked at: LinkedIn, SoFi, Walmart Labs.

Maximum Number of Non-overlapping Palindrome Substrings | LeetCode 2472

Hard

Two Pointers String Dynamic Programming Greedy

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int maxPalindromes(String s, int k) {
        int n = s.length();
        int[] dp = new int[n + 1];

        for (int i = 1; i <= n; i++) {
            dp[i] = dp[i - 1];

            // Check for a palindrome of length k ending at i-1
            if (i >= k) {
                String sub = s.substring(i - k, i);
                String reversedSub = new StringBuilder(sub).reverse().toString();
                if (sub.equals(reversedSub)) {
                    dp[i] = Math.max(dp[i], dp[i - k] + 1);
                }
            }

            // Check for a palindrome of length k+1 ending at i-1
            if (i >= k + 1) {
                String sub = s.substring(i - k - 1, i);
                String reversedSub = new StringBuilder(sub).reverse().toString();
                if (sub.equals(reversedSub)) {
                    dp[i] = Math.max(dp[i], dp[i - k - 1] + 1);
                }
            }
        }

        return dp[n];
    }
}

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Problem Info

Description

Problem 2472: Maximum Number of Non-overlapping Palindrome Substrings - Detailed explanation and solution approaches.

Topics

Two Pointers String Dynamic Programming Greedy

Companies

LinkedInSoFiWalmart Labs

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:2472

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Initialize a DP array `dp` of size `n+1` with all elements set to 0. `dp[i]` will store the maximum number of non-overlapping palindrome substrings in `s[0...i-1]`.
2
Step 2: Iterate through the string `s` from index 1 to `n`. For each index `i`, initialize `dp[i]` to `dp[i-1]`. This represents the case where we don't include any new palindrome ending at `i-1`.
3
Step 3: Check if a palindrome of length `k` exists ending at index `i-1`. Extract the substring `s[i-k:i]` and check if it's a palindrome. If it is, update `dp[i]` to `max(dp[i], dp[i-k] + 1)`.
4
Step 4: Check if a palindrome of length `k+1` exists ending at index `i-1`. Extract the substring `s[i-k-1:i]` and check if it's a palindrome. If it is, update `dp[i]` to `max(dp[i], dp[i-k-1] + 1)`.
5
Step 5: After iterating through the entire string, `dp[n]` will contain the maximum number of non-overlapping palindrome substrings. Return `dp[n]`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the nested loops where the outer loop runs 'n' times and inner string operations take 'k' time. The space complexity arises from storing the 'dp' array.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Greedy Approach

Iterate through the string and greedily pick the longest possible palindrome starting at each position. While this may seem intuitive, it's not guaranteed to produce the optimal result because picking the longest palindrome at each step might prevent us from selecting other valid palindromes later on.

Time: O(n^2)Space: O(1)

Trade-offs: The greedy approach is simpler to implement, but it might not yield the optimal solution. The DP approach guarantees the optimal solution but is slightly more complex.

Manacher's Algorithm

Manacher's algorithm can find all palindrome substrings in O(n) time. We can use it to precompute all palindromes and then use dynamic programming to find the optimal selection of non-overlapping palindromes. Although Manacher's Algorithm computes palindromes faster, dynamic programming for optimal selection remains.

Time: O(n)Space: O(n)

Trade-offs: Manacher's algorithm speeds up palindrome substring detection. However, overall complexity might not decrease compared to the provided DP solution because it is still dependent on the problem statement (selecting optimal count based on length 'k').

✨

Key Insights

1Insight 1: Dynamic programming is a suitable approach because we can build up the solution incrementally. We can store the maximum number of palindromes we can find up to a certain index in the string.
2Insight 2: The core idea is to consider each position `i` in the string. At this position, we can either not include any new palindrome substring ending at `i`, or include a palindrome substring of length `k` or `k+1` (if it exists) ending at `i`. We then take the maximum of these possibilities.
3Insight 3: We only need to check for palindromes of length `k` and `k+1`. Longer palindromes can potentially be broken down into overlapping palindromes of length k and k+1 and not increase our palindrome count, so considering lengths greater than k+1 is redundant.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty string `s`. The DP array is initialized with size `n+1`. The loop doesn't run. The algorithm correctly returns 0, as there are no palindromes.
Edge case 2: `k` is greater than the length of `s`. The inner `if` conditions will prevent the substrings from being accessed. So the algorithm is valid for all cases.
Edge case 3: No palindrome exists in the string. The dp array will maintain its values from initial state (0), and the algorithm returns 0.
Edge case 4: String 's' with length less than 'k'. In this case, the if conditions inside loop will be skipped, dp[i] = dp[i-1], finally dp[n] = 0 since dp[0] = 0. Correct answer since substring length should be at least 'k'.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the non-overlapping constraint correctly. It's crucial to ensure that selected substrings don't overlap.

✗Common mistake 2: Incorrectly checking for palindromes. Ensure the substring is reversed and compared correctly.

✗Common mistake 3: Forgetting to handle the base cases, especially when the string is empty or when no palindrome of length at least `k` exists.

✗Common mistake 4: Not considering palindromes of length k+1, missing a potential better solution.

✗Common mistake 5: Trying to find the longest palindromes first. As highlighted by the greedy solution, this approach is flawed because a shorter palindrome might unlock more valid palindromes later in the string.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Text compression. Finding and replacing palindromic sequences can contribute to reducing the size of text data.

🏦Real-world scenario 2: Bioinformatics. Identifying palindromic sequences in DNA or RNA, which often have significant biological functions (e.g., restriction enzyme recognition sites).

🎮Real-world scenario 3: Pattern recognition in strings. In any application involving sequence analysis, identifying non-overlapping patterns that satisfy certain length constraints can be useful. For example, analyzing network traffic for specific attack signatures, or analyzing financial data for specific trading patterns.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

The two-pointer technique is powerful for sorted arrays. The key insight is recognizing when elements can be processed from both ends simultaneously.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

The most common mistake is not handling the case when pointers meet. Always consider the termination condition.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 11, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Number of Non-overlapping Palindrome Substrings"?

For this hard-level Two Pointers problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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