Given an array of words and a length L, format the text such that each line has exactly L 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 L characters.

Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line do 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.

For example, words: [“This”, “is”, “an”, “example”, “of”, “text”, “justification.”] L: 16.

Return the formatted lines as:

[
"This    is    an",
"example  of text",
"justification.  "
]


Note: Each word is guaranteed not to exceed L in length.

This problem can be solved in two steps in general: (1) count the number of words that can be placed in current line, and then (2) allocate spaces.

For the first step, since the words should be packed in a greedy approach, what we need count is the sum of characters in the $$N$$ words plus the default $$N-1$$ spaces between the N words. Be careful to the last word in this line.

For the second step, we should treat two types of lines differently. Assume there are totally $$M$$ spaces should be inserted in a line. For the lines other than the last line, insert $$M/(N-1)$$ spaces between two nearby words if $$M$$ can be divided by $$N-1$$(where $$N$$ is the number of words in this line). If $$M$$ cannot be divided by $$N-1$$, add 1 space for the first $$M \mod (N-1)$$ words. For the last line, put 1 space between two words, and put other spaces to the end of the last word.

Following is my C++ implementation of text justification:


class Solution {
public:
vector<string> fullJustify(vector<string> &words, int L) {
vector<string> ret;
int word_cnt=0; // count words that have been processed

while(word_cnt < static_cast<int>(words.size())) {
// count words that can be placed in current line
int char_cnt=0; // characters justified in a line
vector<string> vec;
int idx=word_cnt;
while(char_cnt+static_cast<int>(vec.size())-1<=L && idx<static_cast<int>(words.size())) {
char_cnt+=static_cast<int>(words[idx].size());
vec.push_back(words[idx]);
idx++;
}
if(char_cnt+static_cast<int>(vec.size())-1>L) {
char_cnt-=vec.back().size();
vec.pop_back(); // remove the last word because it surpasses the line limit
}
word_cnt+=vec.size();

// assign spaces for this line
string line;
int total_spaces=L-char_cnt;
int avg_spaces; // average spaces for
int left_spaces; //spaces left
if(word_cnt!=static_cast<int>(words.size())) {
if(static_cast<int>(vec.size())>1) { // This line contains more than 1 word
avg_spaces=total_spaces/(static_cast<int>(vec.size())-1);
left_spaces=total_spaces%(static_cast<int>(vec.size())-1);
// Formulate strings per line
for(int i=0;i<static_cast<int>(vec.size())-1;++i) {
line+=vec[i];
for(int j=0;j<avg_spaces;++j)
line+=" ";
if(left_spaces>0) {
line+=" ";
left_spaces--;
}
}
line+=vec.back(); // put the last word to the rightmost position
} else { // Only one word
avg_spaces=0;
left_spaces=total_spaces;
line+=vec.back();
while(left_spaces>0) {
line+=" "; // append spaces to the last word
--left_spaces;
}
}
} else {
// The last line
avg_spaces=1;
left_spaces=total_spaces-static_cast<int>(vec.size())+1;
for(int i=0;i<static_cast<int>(vec.size())-1;++i) {
line+=vec[i];
line+=" ";
}
line+=vec.back();
while(left_spaces>0) {
line+=" "; // append spaces to the last word
--left_spaces;
}
}
ret.push_back(line);
}

return ret;
}
};