68.Text_Justification(Hard).md

68. Text Justification (Hard)

Date and Time: Jul 10, 2024, 14:48 (EST)

Link: https://leetcode.com/problems/text-justification/

Question:

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 \space " instead of "shall \space 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

KeyPoints:

We repeatedly append each word to tmp[] and each word's length to length, if the next word word[i]'s length > maxWidth, we should stop adding this word to current row.

Then, we check how many spaces we need to insert among len(tmp) - 1 elements (len(tmp) is how many elements we added so far) by maxWidth - length, we use (maxWidth - length) // max(1, len(tmp)-1) to know how many spaces to add among each elements in tmp except the last element, and we use max(1, len(tmp)-1) to avoid the case that tmp has only one element, so we want the element to be at least 1. Sometimes it is possible the spaces can't be distributed evenly, so we have remainder = (maxWidth - length) % max(1, len(tmp)-1) as we append to each word and decrement remainder (guaranteed that more spaces on the left).

Next, we reset tmp, length = [], 0 and use "".join(tmp) to make all elements in tmp to be string and append it to res.

Finally, after the for loop is done, we will have one extra line left (as you can see from examples usually the last line can't satisfy if length + len(tmp) + len(words[i]) > maxWidth because we just reset tmp, length). So, we handle it by adding space " ".join(tmp) to put them all together in one line with space in case we have more than one word in the lastLine. Since the last line of text should be left-justified, and no extra space is inserted between words, we should only add extra spaces " " * (maxWidth - len(lastLine)) in the end of lastLine.

My Solution:

class Solution:
    def fullJustify(self, words: List[str], maxWidth: int) -> List[str]:
        # Count how many words can be added to current row within maxWidth
        # keep track of current words length, when a new len(words[i]) > maxWidth, stop adding word
        # calculate the total spaces by maxWidth - total_length
        # each space = total_spaces // max(1, len(tmp) - 1)
        # remainder = total_spaces % max(1, len(tmp) - 1)
        # if remainder, means we have unevenly spaces to assign
        # So we add extra space to each char, then decrement remainder

        # If we have extra words to handle in the last line, we put all the remainder spaces to the right
        
        # TC: O(n), n = len(words), SC: O(n)
        tmp, total_length = [], 0
        res = []
        for i in range(len(words)):
            # When we add the new word length > maxWidth
            if len(words[i]) + total_length + len(tmp) > maxWidth:
                total_spaces = maxWidth - total_length
                spaces = total_spaces // max(1, len(tmp) - 1)
                remainder = total_spaces % max(1, len(tmp) - 1)
                # Traverse words in tmp and add `spaces` to each word
                for j in range(max(1, len(tmp) - 1)):
                    tmp[j] += (" " * spaces)
                    # Add remainder space to each word if we have then decrement
                    if remainder:
                        tmp[j] += " "
                        remainder -= 1
                res.append("".join(tmp))
                tmp, total_length = [], 0   # reset after finish one row
            # Repeatly add words when total_length + words[i] < maxWidth
            total_length += len(words[i])
            tmp.append(words[i])
        # Checking for last line
        lastLine = " ".join(tmp)    # Append a space after each word
        spaces = maxWidth - len(lastLine)
        res.append(lastLine + " " * spaces)     # Append all spaces to the right of lastLine
        return res

Time Complexity: $O(n)$, n = len(words).
Space Complexity: $O(n)$