Text Justification

LeetCode 68 | Difficulty: Hard

Hard

Problem Description

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`

Topics: Array, String, Simulation

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Approach

String Processing

Consider character frequency counts, two-pointer approaches, or building strings efficiently. For pattern matching, think about KMP or rolling hash. For palindromes, expand from center or use DP.

When to use

Anagram detection, palindrome checking, string transformation, pattern matching.

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Solutions

Solution 1: C# (Best: 183 ms)

Metric	Value
Runtime	183 ms
Memory	42.1 MB
Date	2022-02-09

Solution

public class Solution {
    public IList<string> FullJustify(string[] words, int maxWidth) {
      int left = 0;
    List<string> result = new List<string>();
    
    while(left < words.Length)
    {
        int right = FindRight(left, words, maxWidth);
        result.Add(Justify(left, right, words, maxWidth));
        left = right+1;
    }
    return result;

}

private int FindRight(int left, string[] words, int maxWidth)
{
    int right = left;
    int sum = words[right++].Length;
    while(right<words.Length && sum+1+words[right].Length <= maxWidth)
    {
        sum =  sum+1+words[right++].Length;
    }
    return right-1;
}

private string Justify(int left, int right,string[] words, int maxWidth)
{
    if(right-left==0) return PadResult(words[left], maxWidth);
    bool islastLine = right == (words.Length-1);
    int numSpaces = right-left;
    int totalSpace = maxWidth - GetWordsLength(left, right, words);
    
    string space = islastLine ? " " : new string(' ', totalSpace/numSpaces);
    int remainder = islastLine ? 0 : totalSpace%numSpaces;
    
    StringBuilder sb = new StringBuilder();
    for (int i = left; i <= right; i++)
    {
        sb.Append(words[i]).Append(space).Append(remainder-- > 0 ? " " : string.Empty);
    }
    
    return PadResult(sb.ToString().Trim(),maxWidth);
}

private int GetWordsLength(int left, int right, string[] words)
{
    int len = 0;
    for (int i = left; i <= right; i++)
    {
        len += words[i].Length;
    }
    return len;
}

private string PadResult(string result, int maxWidth)
{
    if(maxWidth==result.Length) return result;
    return result+new string(' ',maxWidth-result.Length);
}
}

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Complexity Analysis

Approach	Time	Space
Solution	$O(n)$	$O(1) to O(n)$

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Interview Tips

Key Points

• Break the problem into smaller subproblems. Communicate your approach before coding.