Text Justification
Post
Cancel

# LeetCode #68: Text Justification (C/C++).

hard

source: https://leetcode.com/problems/text-justification/
C/C++ Solution to LeetCode problem 68. Text Justification.

## Problem

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.

## Examples

### Example 1:

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

1 2 3 4 5 [ "This is an", "example of text", "justification. " ] 

### Example 2:

Input: words = [“What”,”must”,”be”,”acknowledgment”,”shall”,”be”], maxWidth = 16
Output:

1 2 3 4 5 [ "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:

1 2 3 4 5 6 7 8 [ "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

## Solution

The approach is easy using greedy algorithm.

• Knowing the maxWidth for a line, we start adding words until we run out of space.
• We calculate the number of spaces abailable for a line (maxWidth - length_of_joint_words).
• The spaces will be divided by the number of words in the line - 1 (for three words, there are two spaces… and so on).
• If we are in the last line (when already iterate through all words), then there is just one space between words, and at the end we add as many spaces as needed to complete the line maxWidth.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 class Solution { public: vector<string> fullJustify(vector<string>& words, int maxWidth) { vector<string> sol; int w=0; int c=0; while (w<words.size()) { int s = 0; while (w < words.size() && (s + 1 + words[w].size()) <= maxWidth + 1) { s += (words[w].size() + 1); w++; } string line = ""s; int maxSpace = maxWidth - (s - (w-c)); while (c < w) { line += words[c]; int nSpaces = 1; bool extra = false; if (w - c == 1) nSpaces = maxWidth - line.size(); else if (w < words.size()) { nSpaces = maxSpace / (w-c-1); extra = (maxSpace % (w-c-1)) != 0; } for (int i=0; i<nSpaces; i++) line += " "s; if (extra) { line += " "s; nSpaces++; } c++; maxSpace -= nSpaces; } sol.push_back(line); } return sol; } };