What is the time complexity of Maximum Deletions on a String?

The optimal solution has a time complexity of O(n^2).

What is the space complexity of Maximum Deletions on a String?

The space complexity is O(n^2).

What data structures are used to solve Maximum Deletions on a String?

This problem uses String, Dynamic Programming, Rolling Hash, String Matching, Hash Function. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Deletions on a String?

Insight 1: Dynamic Programming is suitable because the optimal solution for a prefix of the string depends on the optimal solutions of its subproblems (smaller prefixes). Insight 2: Longest Common Prefix (LCP) array is crucial for efficiently determining if the first `i` letters are equal to the next `i` letters in the string.

Which companies ask LeetCode 2430. Maximum Deletions on a String in interviews?

This problem is frequently asked at DE Shaw.

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Maximum Deletions on a String - LeetCode 2430 Solution

Q: How to solve Maximum Deletions on a String?

The solution uses dynamic programming to determine the maximum number of deletions. First, it computes the longest common prefix (LCP) array `lcp[i][j]`, where `lcp[i][j]` stores the length of the longest common prefix of the substrings starting at indices `i` and `j`. Then, it uses dynamic programming to compute `dp[i]`, which represents the maximum number of deletions starting from index `i`. The final result is `dp[0]`. The LCP array helps in quickly checking if deleting the first `k` characters is a valid operation without comparing the substrings repeatedly.

Maximum Deletions on a String - Complete Solution Guide

Maximum Deletions on a String is LeetCode problem 2430, 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 consisting of only lowercase English letters. In one operation, you can: Delete the entire string s , or Delete the first i letters of s if the first i letters of s are equal to the following i letters in s , for any i in the range 1 <= i <= s.length / 2 . For example, if s = "ababc" , then in one operation, you could delete the first two letters of s to get "abc" , since the first two letters of s and the following two letters of s are both equal to "ab" . Return the ma

Detailed Explanation

The problem asks us to find the maximum number of operations to delete a given string `s` consisting of lowercase English letters. In each operation, we can either delete the entire string, or delete the first `i` characters of `s` if the next `i` characters in `s` are equal to the first `i` characters. The goal is to maximize the number of such operations to completely remove the string.

Solution Approach

The solution uses dynamic programming to determine the maximum number of deletions. First, it computes the longest common prefix (LCP) array `lcp[i][j]`, where `lcp[i][j]` stores the length of the longest common prefix of the substrings starting at indices `i` and `j`. Then, it uses dynamic programming to compute `dp[i]`, which represents the maximum number of deletions starting from index `i`. The final result is `dp[0]`. The LCP array helps in quickly checking if deleting the first `k` characters is a valid operation without comparing the substrings repeatedly.

Step-by-Step Algorithm

Step 1: Calculate the Longest Common Prefix (LCP) array: `lcp[i][j]` stores the length of the longest common prefix between the substrings starting at `i` and `j`. This is done using a nested loop, iterating backwards through the string. If `s[i] == s[j]`, then `lcp[i][j] = lcp[i+1][j+1] + 1`.
Step 2: Initialize the dynamic programming array: `dp[i]` is initialized to 1 for all indices `i`. This is because at the very least, we can delete the entire remaining substring in one operation.
Step 3: Populate the dynamic programming array: Iterate backwards through the string, from `n-1` to `0`. For each index `i`, iterate through possible lengths `k` of the substring to delete, where `k` ranges from 1 to `(n - i) // 2`.
Step 4: Check for valid deletion: If `lcp[i][i + k] >= k`, it means the first `k` characters starting from `i` are equal to the next `k` characters. If this condition is met, update `dp[i]` to `max(dp[i], 1 + dp[i + k])`. This essentially means we are considering the possibility of deleting the first k characters, and adding 1 (the current deletion) to the maximum number of deletions possible from the remaining substring starting at `i+k`.
Step 5: Return the result: After completing the loops, `dp[0]` will contain the maximum number of deletions possible for the entire string.

Key Insights

Insight 1: Dynamic Programming is suitable because the optimal solution for a prefix of the string depends on the optimal solutions of its subproblems (smaller prefixes).
Insight 2: Longest Common Prefix (LCP) array is crucial for efficiently determining if the first `i` letters are equal to the next `i` letters in the string.
Insight 3: The state of the dynamic programming can be defined by the starting position in the string, thus `dp[i]` represents the maximum deletions possible starting from index `i`.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: String, Dynamic Programming, Rolling Hash, String Matching, Hash Function.

Companies

Asked at: DE Shaw.

Maximum Deletions on a String | LeetCode 2430

Hard

String Dynamic Programming Rolling Hash String Matching Hash Function

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

Solution Code

Available in:

class Solution {
    public int deleteString(String s) {
        int n = s.length();
        
        int[][] lcp = new int[n + 1][n + 1];
        for (int i = n - 1; i >= 0; i--) {
            for (int j = i + 1; j < n; j++) {
                if (s.charAt(i) == s.charAt(j)) {
                    lcp[i][j] = 1 + lcp[i + 1][j + 1];
                }
            }
        }
        
        int[] dp = new int[n];
        for (int i = 0; i < n; i++) {
            dp[i] = 1;
        }
        
        for (int i = n - 1; i >= 0; i--) {
            for (int k = 1; k <= (n - i) / 2; k++) {
                if (lcp[i][i + k] >= k) {
                    dp[i] = Math.max(dp[i], 1 + dp[i + k]);
                }
            }
        }
        
        return dp[0];
    }
}

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Problem Info

Description

Problem 2430: Maximum Deletions on a String - Detailed explanation and solution approaches.

Topics

String Dynamic Programming Rolling Hash String Matching Hash Function

Companies

DE Shaw

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2430

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: Calculate the Longest Common Prefix (LCP) array: `lcp[i][j]` stores the length of the longest common prefix between the substrings starting at `i` and `j`. This is done using a nested loop, iterating backwards through the string. If `s[i] == s[j]`, then `lcp[i][j] = lcp[i+1][j+1] + 1`.
2
Step 2: Initialize the dynamic programming array: `dp[i]` is initialized to 1 for all indices `i`. This is because at the very least, we can delete the entire remaining substring in one operation.
3
Step 3: Populate the dynamic programming array: Iterate backwards through the string, from `n-1` to `0`. For each index `i`, iterate through possible lengths `k` of the substring to delete, where `k` ranges from 1 to `(n - i) // 2`.
4
Step 4: Check for valid deletion: If `lcp[i][i + k] >= k`, it means the first `k` characters starting from `i` are equal to the next `k` characters. If this condition is met, update `dp[i]` to `max(dp[i], 1 + dp[i + k])`. This essentially means we are considering the possibility of deleting the first k characters, and adding 1 (the current deletion) to the maximum number of deletions possible from the remaining substring starting at `i+k`.
5
Step 5: Return the result: After completing the loops, `dp[0]` will contain the maximum number of deletions possible for the entire string.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The LCP array calculation takes O(n^2) time due to the nested loops. The dynamic programming calculation also takes O(n^2) time due to the nested loops. Therefore, the overall time complexity is O(n^2). The LCP array requires O(n^2) space and the DP array requires O(n) space. Thus, the overall space complexity is O(n^2).

📖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

Rolling Hash

Instead of calculating LCP using character by character comparison, we can use Rolling Hash to compare substrings in O(1) time (on average). This can potentially reduce the time complexity for checking if the first i characters equal the next i characters.

Time: O(n^2) on average, O(n^3) worst case (due to hash collisions)Space: O(n)

Trade-offs: Rolling Hash can be faster in practice for many inputs, but the worst-case time complexity is higher due to the possibility of hash collisions. Requires careful handling of modulo arithmetic to avoid overflow.

✨

Key Insights

1Insight 1: Dynamic Programming is suitable because the optimal solution for a prefix of the string depends on the optimal solutions of its subproblems (smaller prefixes).
2Insight 2: Longest Common Prefix (LCP) array is crucial for efficiently determining if the first `i` letters are equal to the next `i` letters in the string.
3Insight 3: The state of the dynamic programming can be defined by the starting position in the string, thus `dp[i]` represents the maximum deletions possible starting from index `i`.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

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

Edge Cases

Edge case 1: Empty string (or string of length 1). The code handles the case where string length is 1 correctly by initializing dp[i] to 1 and returning dp[0]
Edge case 2: String with no repeating patterns (e.g., "abcdefg"). The LCP array will have mostly 0s, and the DP array will remain initialized to 1, because no substring will qualify for the deletion condition. `dp[0]` will remain 1.
Edge case 3: String with many repeating patterns (e.g., "aaaaaa"). The LCP array will have large values, and the DP array will be updated at each step leading to the optimal result which would be equivalent to the length of the string since each character can be deleted individually.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the LCP array. Ensuring the loop directions are correct (backwards) and that the base case (`lcp[i+1][j+1]`) is handled properly is crucial.

✗Common mistake 2: Not properly initializing the DP array. The base case for the DP array is that you can always delete the entire string, so each element needs to be initialized to 1.

✗Common mistake 3: Off-by-one errors in indexing when accessing the LCP or DP array.

✅

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: Data compression. Identifying repeating patterns in strings is a key technique used in data compression algorithms like Lempel-Ziv (LZ77).

🏦Real-world scenario 2: Bioinformatics. Identifying repeating sequences in DNA or protein sequences can help in understanding genetic structures and identifying potential drug targets.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

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

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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: August 30, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

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 identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

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 Deletions on a String"?

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

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