**Justify Text - Amazon Top Interview Questions**

### Problem Statement :

Given a list of words and an integer line length k, return a list of strings which represents each line fully justified, with as many words as possible in each line. There should be at least one space between each word, and pad extra spaces when necessary so that each line has exactly length k. Spaces should be distributed as equally as possible, with any extra spaces distributed starting from the left. If you can only fit one word on a line, then pad the right-hand side with spaces. Each word is guaranteed not to be longer than k. Example 1 Input words = ["the", "quick", "brown", "fox", "jumps", "over", "the", "lazy", "dog"] k = 16 Output ["the quick brown", "fox jumps over", "the lazy dog"] Explanation First line: 1 extra space on the left (distributing from the left) Second line: 2 extra spaces distributed evenly Third line: 4 extra spaces distributed evenly

### Solution :

` ````
Solution in C++ :
void fit_in(vector<string>& words, vector<int>& sumsize, int i, int j, int k, vector<string>& ans) {
string wip;
int left_sp = k - sumsize[j + 1] + sumsize[i];
for (int p = i; p < j; p++) {
int sp = left_sp % (j - p) == 0 ? left_sp / (j - p) : 1 + left_sp / (j - p);
left_sp -= sp;
wip.append(words[p]);
wip.append(sp, ' ');
}
wip.append(words[j]);
if (wip.size() < k) wip.append(k - wip.size(), ' ');
ans.push_back(wip);
}
vector<string> solve(vector<string>& words, int k) {
vector<string> ans;
vector<int> sumsize{0};
for (int i = 0, s = 0; i < words.size(); i++) {
s += words[i].size();
sumsize.push_back(s);
}
for (int i = 0, j = 0; i < words.size();) {
while (j < words.size() && sumsize[j + 1] - sumsize[i] + j - i <= k) j++;
fit_in(words, sumsize, i, j - 1, k, ans);
i = j;
}
return ans;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public String[] solve(String[] words, int k) {
List<String> res = new ArrayList<>();
int len = words.length;
StringBuilder sb = new StringBuilder();
int index = 0;
while (index < len) {
String curStr = words[index];
int curLen = curStr.length();
if (curLen + sb.length() > k) {
sb = sb.deleteCharAt(sb.length() - 1);
StringBuilder evenSb = doEvenDirstributed(sb, k);
System.out.println("evenSb" + evenSb.toString());
sb = new StringBuilder();
res.add(evenSb.toString());
continue;
}
sb.append(curStr).append(" ");
System.out.println("Sb" + sb.toString());
index++;
}
StringBuilder lastEvenSb = doEvenDirstributed(sb, k);
res.add(lastEvenSb.toString());
String[] result = new String[res.size()];
for (int i = 0; i < res.size(); i++) {
result[i] = res.get(i);
}
return result;
}
private StringBuilder doEvenDirstributed(StringBuilder sb, int k) {
String str = sb.toString().trim();
String[] stringArr = str.split(" ");
int len = sb.length();
int letterCount = 0;
for (String s : stringArr) {
letterCount += s.length();
}
int needSpaceCount = k - letterCount;
int x = 1;
int y = 0;
if (stringArr.length - 1 == 0) {
x = k - stringArr[0].length();
y = 0;
} else {
x = needSpaceCount / (stringArr.length - 1);
y = needSpaceCount % (stringArr.length - 1);
}
System.out.println(
"need:" + needSpaceCount + " x:" + x + " y:" + y + "letterCount" + letterCount);
StringBuilder resStringBuilder = new StringBuilder();
for (int i = 0; i < stringArr.length; i++) {
resStringBuilder.append(stringArr[i]);
if (i != 0 && i == stringArr.length - 1) {
continue;
}
int xTemp = x;
int yTemp = y;
int c = 0;
while (c < xTemp) {
resStringBuilder.append(" ");
c++;
}
if (y != 0) {
resStringBuilder.append(" ");
y--;
}
}
return resStringBuilder;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, words, k):
def round_robin(line, count):
"""
round robin addition of spaces starting from left to right
until we reach the length of string = k
"""
for i in range(k - count):
line[i % (len(line) - 1 or 1)] += " "
return "".join(line)
ans, current_line, current_line_length = [], [], 0
for word in words:
# if addition of this word makes the current line length > k,
# we justify the current line, add to ans and reset the current line
# we add len(current_line) since it's the number of spaces after each word
if current_line_length + len(word) + len(current_line) > k:
ans.append(round_robin(current_line, current_line_length))
current_line, current_line_length = [], 0
# else just add current word and its length to the current line and its length
current_line.append(word)
current_line_length += len(word)
ans.append(round_robin(current_line, current_line_length))
return ans
```

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