LeetCode - #68 Align text left and right

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Difficulty Level: Difficult

1. Description

Given an array of words wordsand a length maxWidth, rearrange the words to be text with exactly characters maxWidthper , justified left and right.

You should use a " greedy algorithm " to place a given word; that is, put as many words as possible on each line. It can be ' 'padded so that each line has exactly maxWidth characters.

It is required to distribute the number of spaces between words as evenly as possible. If the spaces between words on a line are not evenly distributed, place more spaces on the left than on the right.

The last line of text should be left-aligned with no extra spaces inserted between words.

Notice:

• A word is a sequence of characters consisting of non-space characters.
• The length of each word is greater than 0 and less than or equal to maxWidth .
• The input word array contains wordsat least one word.

2. Example

Example 1

输入: words = ["This", "is", "an", "example", "of", "text", "justification."], maxWidth = 16
输出:
[
   "This    is    an",
   "example  of text",
   "justification.  "
]
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Example 2

输入:words = ["What","must","be","acknowledgment","shall","be"], maxWidth = 16
输出:
[
  "What   must   be",
  "acknowledgment  ",
  "shall be        "
]
解释: 注意最后一行的格式应为 "shall be    " 而不是 "shall     be",
     因为最后一行应为左对齐，而不是左右两端对齐。       
     第二行同样为左对齐，这是因为这行只包含一个单词。
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Example 3

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

• 1 <= words.length <= 300
• 1 <= words[i].length <= 20
• words[i]Consists of lowercase English letters and symbols
• 1 <= maxWidth <= 100
• words[i].length <= maxWidth

3. Answers

class TextJustification {
    func fullJustify(_ words: [String], _ maxWidth: Int) -> [String] {
        var res = [String]()
        var last = 0, currentLineLength = 0
        
        for (i, word) in words.enumerated() {
            if currentLineLength + word.count + (i - last) > maxWidth {
                
                res.append(buildLine(words, last, i - 1, maxWidth, currentLineLength))
                
                last = i
                currentLineLength = 0   
            }
            
            currentLineLength += word.count
        }
        
        res.append(buildLastLine(words, last, words.count - 1, maxWidth))
        
        return res
    }
    
    fileprivate func buildLine(_ words: [String], _ start: Int, _ end: Int, _ maxWidth: Int, _ currentLineLength: Int) -> String {
        var line = ""
        var extraSpaceNum = 0, spaceNum = 0
        
        if end > start {
            extraSpaceNum = (maxWidth - currentLineLength) % (end - start)
            spaceNum = (maxWidth - currentLineLength) / (end - start)
        } else {
            spaceNum = maxWidth - currentLineLength
        }
        
        for i in start...end {
            line.append(words[i])
            
            if start != end && i == end {
                break
            } 
            
            for _ in 0..<spaceNum {
                line.append(" ")
            }
            
            if extraSpaceNum > 0 {
                line.append(" ")
                extraSpaceNum -= 1
            }
        }
        
        return line
    }
    
    fileprivate func buildLastLine(_ words: [String], _ start: Int, _ end: Int, _ maxWidth: Int) -> String {
        var line = ""
        
        for i in start...end {
            line.append(words[i])
            
            if i < end {
                line.append(" ")
            }
        }
        
        while line.count < maxWidth {
            line.append(" ")
        }
        
        return line
    }
}
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• Main idea: Iterate over words, keeping track of the index of the first word and the length of the line. Insert spaces with fixed spaces and extra spaces. .
• Time Complexity: O(n)
• Space Complexity: O(n)

Repository for the algorithm solution: LeetCode-Swift

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