What is the time complexity of Text Justification?

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

What data structures are used to solve Text Justification?

This problem uses Array, String, Simulation. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Text Justification?

Insight 1: The core task is to determine how many words fit on each line before exceeding the maxWidth. This involves iterating through the words and accumulating their lengths along with the spaces between them. Insight 2: Distributing spaces evenly between words in a line requires calculating the base number of spaces and any remaining extra spaces. The extra spaces need to be distributed from left to right.

Which companies ask LeetCode 68. Text Justification in interviews?

This problem is frequently asked at Adobe, Airbnb, Amazon, Apple, Atlassian and 28 more companies.

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

Text Justification - LeetCode 68 Solution

Text Justification - Complete Solution Guide

Text Justification is LeetCode problem 68, 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

Given an array of strings words and a width maxWidth , format the text such that each line has exactly maxWidth characters and is fully (left and right) justified. You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad extra spaces ' ' when necessary so that each line has exactly maxWidth characters. Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line does not divide evenly between words, the

Detailed Explanation

The Text Justification problem requires formatting a given array of words into lines of text, each with a specified maximum width. The words must be packed greedily, meaning as many words as possible are placed on each line. The lines, except for the last one, should be fully justified (left and right). This involves distributing extra spaces evenly between words. If spaces can't be distributed evenly, the leftmost gaps should have more spaces. The last line must be left-justified with no extra spaces inserted between words.

Solution Approach

The provided solution uses a greedy approach to pack words into lines. It iterates through the words array, building each line one at a time. For each line, it determines how many words can fit within the maxWidth. Then, based on whether it's the last line or a line with a single word, it applies different justification rules. The solution calculates the required spaces, distributes them evenly where applicable, and constructs the final justified string for each line.

Step-by-Step Algorithm

Step 1: Initialize an empty list `result` to store the justified lines.
Step 2: Iterate through the `words` array using a `while` loop.
Step 3: For each line, determine the ending index `j` by greedily adding words until the `maxWidth` is exceeded.
Step 4: Extract the words for the current line `line_words` from `i` to `j`.
Step 5: Check if it is the last line or if `line_words` contains only one word.
Step 6: If it is the last line or only one word, left-justify the line and add it to `result`.
Step 7: If it is not the last line and contains more than one word, calculate the required number of spaces.
Step 8: Distribute spaces evenly and add extra spaces from left to right.
Step 9: Build the justified line using a string builder and append it to the `result`.
Step 10: Update the starting index `i` to `j` and repeat until all words are processed.

Key Insights

Insight 1: The core task is to determine how many words fit on each line before exceeding the maxWidth. This involves iterating through the words and accumulating their lengths along with the spaces between them.
Insight 2: Distributing spaces evenly between words in a line requires calculating the base number of spaces and any remaining extra spaces. The extra spaces need to be distributed from left to right.
Insight 3: Handling the last line differently (left-justified) is a crucial edge case. Also, lines with only one word must be left-justified.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, String, Simulation.

Companies

Asked at: Adobe, Airbnb, Amazon, Apple, Atlassian, Bloomberg, Capital One, Coinbase, Coursera, Databricks, DevRev, Google, Karat, LinkedIn, Meta, Microsoft, MongoDB, Moveworks, Netflix, Notion, Oracle, PayPal, Robinhood, Roblox, SIG, Samsara, Sentry, TikTok, Twilio, Uber, Visa, WeRide, Zoho.

Text Justification | LeetCode 68

Hard

Array String Simulation

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<String> fullJustify(String[] words, int maxWidth) {
        List<String> result = new ArrayList<>();
        int i = 0;
        int n = words.length;

        while (i < n) {
            int j = i + 1;
            int current_line_words_len = words[i].length();
            while (j < n && (current_line_words_len + words[j].length() + (j - i)) <= maxWidth) {
                current_line_words_len += words[j].length();
                j += 1;
            }

            List<String> line_words = new ArrayList<>();
            for (int k = i; k < j; k++) {
                line_words.add(words[k]);
            }
            int num_words = line_words.size();
            boolean is_last_line = (j == n);

            if (is_last_line || num_words == 1) {
                String line_str = String.join(" ", line_words);
                while (line_str.length() < maxWidth) {
                    line_str += " ";
                }
                result.add(line_str);
            } else {
                int num_gaps = num_words - 1;
                int total_spaces = maxWidth - current_line_words_len;

                int base_spaces = total_spaces / num_gaps;
                int extra_spaces = total_spaces % num_gaps;

                StringBuilder line_builder = new StringBuilder(line_words.get(0));
                for (int k = 1; k < num_words; k++) {
                    int spaces_to_add = base_spaces + (k - 1 < extra_spaces ? 1 : 0);
                    for (int l = 0; l < spaces_to_add; l++) {
                        line_builder.append(' ');
                    }
                    line_builder.append(line_words.get(k));
                }
                result.add(line_builder.toString());
            }

            i = j;
        }

        return result;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 68: Text Justification - Detailed explanation and solution approaches.

Topics

Array String Simulation

Companies

AdobeAirbnbAmazonAppleAtlassianBloombergCapital OneCoinbase

Stats

Difficulty:Hard

Languages:4

Companies:33

Problem #:68

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

10 steps

1
Step 1: Initialize an empty list `result` to store the justified lines.
2
Step 2: Iterate through the `words` array using a `while` loop.
3
Step 3: For each line, determine the ending index `j` by greedily adding words until the `maxWidth` is exceeded.
4
Step 4: Extract the words for the current line `line_words` from `i` to `j`.
5
Step 5: Check if it is the last line or if `line_words` contains only one word.
6
Step 6: If it is the last line or only one word, left-justify the line and add it to `result`.
7
Step 7: If it is not the last line and contains more than one word, calculate the required number of spaces.
8
Step 8: Distribute spaces evenly and add extra spaces from left to right.
9
Step 9: Build the justified line using a string builder and append it to the `result`.
10
Step 10: Update the starting index `i` to `j` and repeat until all words are processed.

💡

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 O(n) because we iterate through the words array at most twice. Space complexity is O(n) because the output list can contain at most 'n' strings, one for each word, and the string manipulations involve allocating space at most proportional to the number of input words.

📖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

Dynamic Programming

A dynamic programming approach could precompute the cost of placing different combinations of words on each line. This can be useful to optimize the 'beauty' of the text, but doesn't align with the problem specification's 'greedy' requirement.

Time: O(n^2) or O(n^3) depending on optimization.Space: O(n)

Trade-offs: DP is overkill here, as the problem specifically requests a greedy approach. DP would add unnecessary complexity without providing significant benefit in this specific scenario.

✨

Key Insights

1Insight 1: The core task is to determine how many words fit on each line before exceeding the maxWidth. This involves iterating through the words and accumulating their lengths along with the spaces between them.
2Insight 2: Distributing spaces evenly between words in a line requires calculating the base number of spaces and any remaining extra spaces. The extra spaces need to be distributed from left to right.
3Insight 3: Handling the last line differently (left-justified) is a crucial edge case. Also, lines with only one word must be left-justified.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

String

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

Edge Cases

Edge case 1: Last line. The last line must be left-justified. The solution explicitly checks for this condition (`j == n`) and applies left justification logic using `ljust()` in Python or a similar approach in other languages.
Edge case 2: Line with only one word. A line containing only one word must also be left-justified. The solution correctly handles this condition by combining it with the last line condition `(is_last_line or num_words == 1)`.
Edge case 3: Words with length equal to maxWidth. This is handled correctly as only a single word will form the line in this case.
Edge case 4: maxWidth is small and most words can't be on the same line. Solution still works as it adds words greedily to the line.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect space distribution. Failing to distribute extra spaces correctly, especially when the number of spaces does not divide evenly between words, is a frequent mistake.

✗Common mistake 2: Handling the last line justification incorrectly. Not treating the last line differently and fully justifying it like the other lines leads to wrong output.

✗Common mistake 3: Off-by-one errors in calculating spaces or lengths. Careful attention is needed to calculate the number of spaces between words and to correctly track the lengths of words and lines.

✗Common mistake 4: Memory management issues in C. Forgetting to allocate or free memory properly in C can lead to memory leaks or segmentation faults.

✅

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 editors and word processors. These applications use text justification algorithms to format documents and create visually appealing layouts.

🏦Real-world scenario 2: Web page layout. Web browsers use similar techniques to format text on web pages, ensuring consistent alignment and readability.

🎮Real-world scenario 3: E-readers and digital publishing. Justification algorithms are used to format text in e-books, magazines, and other digital publications.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

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

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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 8, 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

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

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 "Text Justification"?

For this hard-level Array 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.

🤖 Need help with this problem?