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

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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

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• 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

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The approach is easy using greedy algorithm.

• Knowing the maxWidth for a line, we start adding words until we run out of space.
  • For each added word, we add an exrea space.
• Once we finish with the line, when there is no more space for other word:
  • 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 the previous step gives a fraction of space, we will add an extra espace.
  • After adding a word and the spaces, we reduce the amount of spaces abailable for the line.
  • We calculate again the espaces for the next word.
• 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.

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;
  }
};