What is the time complexity of Split Message Based on Limit?

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

What data structures are used to solve Split Message Based on Limit?

This problem uses String, Binary Search, Enumeration. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Split Message Based on Limit?

Insight 1: The total number of parts ('b') significantly impacts the available space for the message content in each part because the suffix ' ''s length depends on both 'a' and 'b'. Insight 2: To minimize the number of parts, we need to iterate over possible values of 'b' and check if the available space across all parts is sufficient to contain the entire message.

Which companies ask LeetCode 2468. Split Message Based on Limit in interviews?

This problem is frequently asked at Capital One, Databricks, Roblox, Visa, ZipRecruiter.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Split Message Based on Limit - LeetCode 2468 Solution

Split Message Based on Limit - Complete Solution Guide

Split Message Based on Limit is LeetCode problem 2468, 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, message , and a positive integer, limit . You must split message into one or more parts based on limit . Each resulting part should have the suffix "<a/b>" , where "b" is to be replaced with the total number of parts and "a" is to be replaced with the index of the part, starting from 1 and going up to b . Additionally, the length of each resulting part (including its suffix) should be equal to limit , except for the last part whose length can be at most limit . The result

Detailed Explanation

The problem requires splitting a given message into multiple parts, each having a maximum length defined by 'limit'. Each part must include a suffix of the form '<a/b>', where 'a' is the part number (starting from 1) and 'b' is the total number of parts. The goal is to find the minimum number of parts ('b') needed to split the message while adhering to the length limit and suffix constraints. If no such split is possible, return an empty array. The concatenated message parts (without suffixes) must reconstruct the original message.

Solution Approach

The solution uses an iterative approach to find the minimum number of parts ('b'). It iterates through possible values of 'b' (from 1 up to message length + 1) and calculates the total available space for message content based on the given 'limit' and the length of the suffixes '<a/b>'. For each 'b', it checks if the total available space is sufficient to accommodate the entire message. Once a suitable 'b' is found, the message is split into 'b' parts, each with the appropriate suffix.

Step-by-Step Algorithm

Step 1: Initialize 'b' to -1 (indicating no solution found yet).
Step 2: Iterate through possible values of 'current_b' from 1 to message length + 1.
Step 3: For each 'current_b', calculate the length of the number 'current_b' (len_b).
Step 4: Check if the limit is sufficient to include at least one character of message and the suffix. If the limit is less than or equal to the maximum possible suffix length (3 + 2 * len_b), continue to the next 'current_b'.
Step 5: Calculate the total length of all suffix indices (sum of length of string representation of numbers 1 to current_b).
Step 6: Calculate the total length occupied by all suffixes (total_suffix_len) based on 'current_b' and len_b.
Step 7: Calculate the total content capacity (content_capacity) which is the total available space for the actual message across all the parts.
Step 8: If the message length is less than or equal to the content capacity, set 'b' to 'current_b' and break the loop.
Step 9: If 'b' remains -1 after the loop, it means no suitable 'b' was found, so return an empty array.
Step 10: If a suitable 'b' was found, create an array of strings to store the resulting parts.
Step 11: Iterate from a = 1 to 'b' and for each 'a', calculate the length of the suffix and the payload length (available space for the actual message).
Step 12: Extract the appropriate portion of the message (part_message) and create the part string by concatenating 'part_message' and the suffix '<a/b>'.
Step 13: Store the part string in the result array and update the message index.
Step 14: Return the result array.

Key Insights

Insight 1: The total number of parts ('b') significantly impacts the available space for the message content in each part because the suffix '<a/b>''s length depends on both 'a' and 'b'.
Insight 2: To minimize the number of parts, we need to iterate over possible values of 'b' and check if the available space across all parts is sufficient to contain the entire message.
Insight 3: The problem can be viewed as finding the smallest 'b' that satisfies the capacity constraint: total available space for content >= length of the message.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: String, Binary Search, Enumeration.

Companies

Asked at: Capital One, Databricks, Roblox, Visa, ZipRecruiter.

Split Message Based on Limit | LeetCode 2468

Hard

String Binary Search Enumeration

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

Solution Code

Available in:

class Solution {
    public String[] splitMessage(String message, int limit) {
        int n = message.length();
        
        int b = -1;
        
        int total_len_of_indices = 0;
        for (int current_b = 1; current_b <= n + 1; ++current_b) {
            if (current_b > n && b != -1) { // Optimization
                break;
            }
            
            int len_b = String.valueOf(current_b).length();
            
            // The length of the longest possible suffix for 'current_b' parts.
            // This occurs for part 'current_b' itself.
            // Suffix is "<current_b/current_b>", length is 3 + len_b + len_b.
            // The limit must be greater than this to allow for at least one character of message.
            // Since payload can be zero, we need limit > suffix length.
            int max_suffix_len = 3 + 2 * len_b;
            if (limit <= max_suffix_len) {
                continue;
            }
            
            // total_len_of_indices accumulates sum(len(str(i))) for i=1 to current_b
            total_len_of_indices += len_b;
            
            // Total length of all suffixes if we have 'current_b' parts
            // For each part 'a', suffix is "<a/b>". Length is 3 + len(str(a)) + len(str(b))
            // Sum for a=1..b:  b * (3 + len_b) + sum(len(str(a)) for a=1..b)
            int total_suffix_len = current_b * (3 + len_b) + total_len_of_indices;
            
            // Total available space for message content
            // sum(limit - suffix_len(a, b) for a=1..b)
            // = b * limit - total_suffix_len
            int content_capacity = current_b * limit - total_suffix_len;
            
            if (n <= content_capacity) {
                b = current_b;
                break;
            }
        }

        if (b == -1) {
            return new String[0];
        }

        String[] result = new String[b];
        int message_idx = 0;
        int len_b_final = String.valueOf(b).length();
        for (int a = 1; a <= b; ++a) {
            int len_a = String.valueOf(a).length();
            int suffix_len = 3 + len_a + len_b_final;
            int payload_len = limit - suffix_len;
            
            String part_message = message.substring(message_idx, Math.min(message_idx + payload_len, message.length()));
            result[a - 1] = part_message + "<" + a + "/" + b + ">";
            message_idx += payload_len;
        }
        
        return result;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2468: Split Message Based on Limit - Detailed explanation and solution approaches.

Topics

String Binary Search Enumeration

Companies

Capital OneDatabricksRobloxVisaZipRecruiter

Stats

Difficulty:Hard

Languages:4

Companies:5

Problem #:2468

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

14 steps

1
Step 1: Initialize 'b' to -1 (indicating no solution found yet).
2
Step 2: Iterate through possible values of 'current_b' from 1 to message length + 1.
3
Step 3: For each 'current_b', calculate the length of the number 'current_b' (len_b).
4
Step 4: Check if the limit is sufficient to include at least one character of message and the suffix. If the limit is less than or equal to the maximum possible suffix length (3 + 2 * len_b), continue to the next 'current_b'.
5
Step 5: Calculate the total length of all suffix indices (sum of length of string representation of numbers 1 to current_b).
6
Step 6: Calculate the total length occupied by all suffixes (total_suffix_len) based on 'current_b' and len_b.
7
Step 7: Calculate the total content capacity (content_capacity) which is the total available space for the actual message across all the parts.
8
Step 8: If the message length is less than or equal to the content capacity, set 'b' to 'current_b' and break the loop.
9
Step 9: If 'b' remains -1 after the loop, it means no suitable 'b' was found, so return an empty array.
10
Step 10: If a suitable 'b' was found, create an array of strings to store the resulting parts.
11
Step 11: Iterate from a = 1 to 'b' and for each 'a', calculate the length of the suffix and the payload length (available space for the actual message).
12
Step 12: Extract the appropriate portion of the message (part_message) and create the part string by concatenating 'part_message' and the suffix '<a/b>'.
13
Step 13: Store the part string in the result array and update the message index.
14
Step 14: Return the result array.

💡

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)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the nested loop structure implied by the iterative search for 'b' and the string operations within the loop. The space complexity is primarily influenced by the size of the output array and the size of strings.

📖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

Binary Search

Binary search could be employed to optimize the search for the minimum value of 'b' instead of linear iteration. The lower bound for 'b' is 1, and the upper bound is 'n' (message length). We can then use a binary search to find the smallest 'b' that satisfies the condition content_capacity >= n.

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

Trade-offs: Binary search can reduce the time complexity, especially when message length is large. The tradeoff is added code complexity of binary search, however it is still better in terms of performance as binary search provides logarithmic search.

✨

Key Insights

1Insight 1: The total number of parts ('b') significantly impacts the available space for the message content in each part because the suffix '<a/b>''s length depends on both 'a' and 'b'.
2Insight 2: To minimize the number of parts, we need to iterate over possible values of 'b' and check if the available space across all parts is sufficient to contain the entire message.
3Insight 3: The problem can be viewed as finding the smallest 'b' that satisfies the capacity constraint: total available space for content >= length of the message.

📚

Key Terms Explained

For Students

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

String

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

Edge Cases

Edge case 1: When the limit is too small to accommodate even a single character of the message along with the suffix. The solution handles this by checking if 'limit' <= (3 + 2 * len_b) and continuing to the next iteration if it is. This ensures that an empty array is returned when no split is possible.
Edge case 2: When the message is very short and can be split into a small number of parts. The algorithm correctly calculates the number of parts ('b') and splits the message accordingly.
Edge case 3: Message consists of only spaces. The algorithm handles this case correctly, splitting spaces into different parts when necessary to satisfy length constraints.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the available space due to errors in calculating suffix lengths or index lengths.

✗Common mistake 2: Not handling the edge case where the limit is too small to accommodate even one character of the message with its suffix.

✗Common mistake 3: Incorrectly forming the substrings or concatenating the suffixes.

✗Common mistake 4: Using integer overflow, when calculating total suffix length or content capacity especially in C/C++.

✅

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: Sending long SMS messages by splitting them into multiple parts, each with a sequence number.

🏦Real-world scenario 2: Breaking down large data files into smaller chunks for transmission over a network, where each chunk has a header containing information about its position in the file.

🎮Real-world scenario 3: Pagination of results on a web page, where each page displays a limited number of items and includes navigation links to other pages.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

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

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 7, 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

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

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 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 avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

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 "Split Message Based on Limit"?

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 binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?