LeetCode #68 — HARD

Text Justification

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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 empty slots on the left will be assigned more spaces than the slots on the right.

For the last line of text, it should be left-justified, and no extra space is inserted between words.

Note:

A word is defined as a character sequence consisting of non-space characters only.
Each word's length is guaranteed to be greater than 0 and not exceed maxWidth.
The input array words contains at least one word.

Example 1:

Input: words = ["This", "is", "an", "example", "of", "text", "justification."], maxWidth = 16
Output:
[
   "This    is    an",
   "example  of text",
   "justification.  "
]

Example 2:

Input: words = ["What","must","be","acknowledgment","shall","be"], maxWidth = 16
Output:
[
  "What   must   be",
  "acknowledgment  ",
  "shall be        "
]
Explanation: Note that the last line is "shall be    " instead of "shall     be", because the last line must be left-justified instead of fully-justified.
Note that the second line is also left-justified because it contains only one word.

Example 3:

Input: words = ["Science","is","what","we","understand","well","enough","to","explain","to","a","computer.","Art","is","everything","else","we","do"], maxWidth = 20
Output:
[
  "Science  is  what we",
  "understand      well",
  "enough to explain to",
  "a  computer.  Art is",
  "everything  else  we",
  "do                  "
]

Constraints:

1 <= words.length <= 300
1 <= words[i].length <= 20
words[i] consists of only English letters and symbols.
1 <= maxWidth <= 100
words[i].length <= maxWidth

Step 01

Brute Force Baseline

Problem summary: 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 empty slots on the left will be assigned more spaces than the slots on the right. For the last line of text, it should be left-justified, and no extra space is inserted between words. Note: A word is defined as a character sequence consisting of non-space characters only. Each word's length is guaranteed to be greater than 0 and not exceed maxWidth. The input array words contains at least one

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

["This", "is", "an", "example", "of", "text", "justification."]
16

Example 2

["What","must","be","acknowledgment","shall","be"]
16

Example 3

["Science","is","what","we","understand","well","enough","to","explain","to","a","computer.","Art","is","everything","else","we","do"]
20

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

import java.util.*;

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

        while (i < words.length) {
            int j = i + 1;
            int lineChars = words[i].length();

            // Greedily pack as many words as possible into the line.
            while (j < words.length && lineChars + 1 + words[j].length() <= maxWidth) {
                lineChars += 1 + words[j].length();
                j++;
            }

            int wordCount = j - i;
            StringBuilder line = new StringBuilder();

            if (j == words.length || wordCount == 1) {
                // Last line (or single-word line): left-justified.
                for (int k = i; k < j; k++) {
                    if (k > i) line.append(' ');
                    line.append(words[k]);
                }
                while (line.length() < maxWidth) line.append(' ');
            } else {
                int totalWordChars = 0;
                for (int k = i; k < j; k++) totalWordChars += words[k].length();
                int spaces = maxWidth - totalWordChars;
                int gaps = wordCount - 1;
                int even = spaces / gaps;
                int extra = spaces % gaps;

                for (int k = i; k < j - 1; k++) {
                    line.append(words[k]);
                    int count = even + ((k - i) < extra ? 1 : 0);
                    for (int s = 0; s < count; s++) line.append(' ');
                }
                line.append(words[j - 1]);
            }

            ans.add(line.toString());
            i = j;
        }

        return ans;
    }
}

import "strings"

func fullJustify(words []string, maxWidth int) []string {
    ans := []string{}
    i := 0

    for i < len(words) {
        j := i + 1
        lineChars := len(words[i])
        for j < len(words) && lineChars+1+len(words[j]) <= maxWidth {
            lineChars += 1 + len(words[j])
            j++
        }

        wordCount := j - i
        var line strings.Builder

        if j == len(words) || wordCount == 1 {
            for k := i; k < j; k++ {
                if k > i {
                    line.WriteByte(' ')
                }
                line.WriteString(words[k])
            }
            for line.Len() < maxWidth {
                line.WriteByte(' ')
            }
        } else {
            totalWordChars := 0
            for k := i; k < j; k++ {
                totalWordChars += len(words[k])
            }
            spaces := maxWidth - totalWordChars
            gaps := wordCount - 1
            even := spaces / gaps
            extra := spaces % gaps

            for k := i; k < j-1; k++ {
                line.WriteString(words[k])
                count := even
                if (k - i) < extra {
                    count++
                }
                line.WriteString(strings.Repeat(" ", count))
            }
            line.WriteString(words[j-1])
        }

        ans = append(ans, line.String())
        i = j
    }

    return ans
}

class Solution:
    def fullJustify(self, words: List[str], maxWidth: int) -> List[str]:
        ans: List[str] = []
        i = 0

        while i < len(words):
            j = i + 1
            line_chars = len(words[i])
            while j < len(words) and line_chars + 1 + len(words[j]) <= maxWidth:
                line_chars += 1 + len(words[j])
                j += 1

            line_words = words[i:j]
            word_count = len(line_words)

            if j == len(words) or word_count == 1:
                line = ' '.join(line_words)
                line += ' ' * (maxWidth - len(line))
            else:
                total_word_chars = sum(len(w) for w in line_words)
                spaces = maxWidth - total_word_chars
                gaps = word_count - 1
                even, extra = divmod(spaces, gaps)

                parts = []
                for idx, w in enumerate(line_words[:-1]):
                    parts.append(w)
                    parts.append(' ' * (even + (1 if idx < extra else 0)))
                parts.append(line_words[-1])
                line = ''.join(parts)

            ans.append(line)
            i = j

        return ans

impl Solution {
    pub fn full_justify(words: Vec<String>, max_width: i32) -> Vec<String> {
        let max_width = max_width as usize;
        let mut ans = Vec::new();
        let mut i = 0usize;

        while i < words.len() {
            let mut j = i + 1;
            let mut line_chars = words[i].len();
            while j < words.len() && line_chars + 1 + words[j].len() <= max_width {
                line_chars += 1 + words[j].len();
                j += 1;
            }

            let word_count = j - i;
            let mut line = String::new();

            if j == words.len() || word_count == 1 {
                for k in i..j {
                    if k > i {
                        line.push(' ');
                    }
                    line.push_str(&words[k]);
                }
                line.push_str(&" ".repeat(max_width - line.len()));
            } else {
                let mut total_word_chars = 0usize;
                for k in i..j {
                    total_word_chars += words[k].len();
                }

                let spaces = max_width - total_word_chars;
                let gaps = word_count - 1;
                let even = spaces / gaps;
                let extra = spaces % gaps;

                for k in i..j - 1 {
                    line.push_str(&words[k]);
                    let count = even + if (k - i) < extra { 1 } else { 0 };
                    line.push_str(&" ".repeat(count));
                }
                line.push_str(&words[j - 1]);
            }

            ans.push(line);
            i = j;
        }

        ans
    }
}

function fullJustify(words: string[], maxWidth: number): string[] {
  const ans: string[] = [];
  let i = 0;

  while (i < words.length) {
    let j = i + 1;
    let lineChars = words[i].length;
    while (j < words.length && lineChars + 1 + words[j].length <= maxWidth) {
      lineChars += 1 + words[j].length;
      j++;
    }

    const wordCount = j - i;
    let line = '';

    if (j === words.length || wordCount === 1) {
      line = words.slice(i, j).join(' ');
      line += ' '.repeat(maxWidth - line.length);
    } else {
      let totalWordChars = 0;
      for (let k = i; k < j; k++) totalWordChars += words[k].length;
      const spaces = maxWidth - totalWordChars;
      const gaps = wordCount - 1;
      const even = Math.floor(spaces / gaps);
      const extra = spaces % gaps;

      for (let k = i; k < j - 1; k++) {
        line += words[k];
        line += ' '.repeat(even + (k - i < extra ? 1 : 0));
      }
      line += words[j - 1];
    }

    ans.push(line);
    i = j;
  }

  return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(L)

Space

O(L)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.