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Text Justification – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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
0and not exceedmaxWidth. - The input array
wordscontains 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 " 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:
1 <= words.length <= 3001 <= words[i].length <= 20words[i]consists of only English letters and symbols.1 <= maxWidth <= 100words[i].length <= maxWidth
C++ Text Justification LeetCode Solution
vector<string> fullJustify(vector<string> &words, int L) {
vector<string> res;
for(int i = 0, k, l; i < words.size(); i += k) {
for(k = l = 0; i + k < words.size() and l + words[i+k].size() <= L - k; k++) {
l += words[i+k].size();
}
string tmp = words[i];
for(int j = 0; j < k - 1; j++) {
if(i + k >= words.size()) tmp += " ";
else tmp += string((L - l) / (k - 1) + (j < (L - l) % (k - 1)), ' ');
tmp += words[i+j+1];
}
tmp += string(L - tmp.size(), ' ');
res.push_back(tmp);
}
return res;
}
Java Text Justification LeetCode Solution
public List<String> fullJustify(String[] words, int maxWidth) {
int left = 0; List<String> result = new ArrayList<>();
while (left < words.length) {
int right = findRight(left, words, maxWidth);
result.add(justify(left, right, words, maxWidth));
left = right + 1;
}
return result;
}
private int findRight(int left, String[] words, int maxWidth) {
int right = left;
int sum = words[right++].length();
while (right < words.length && (sum + 1 + words[right].length()) <= maxWidth)
sum += 1 + words[right++].length();
return right - 1;
}
private String justify(int left, int right, String[] words, int maxWidth) {
if (right - left == 0) return padResult(words[left], maxWidth);
boolean isLastLine = right == words.length - 1;
int numSpaces = right - left;
int totalSpace = maxWidth - wordsLength(left, right, words);
String space = isLastLine ? " " : blank(totalSpace / numSpaces);
int remainder = isLastLine ? 0 : totalSpace % numSpaces;
StringBuilder result = new StringBuilder();
for (int i = left; i <= right; i++)
result.append(words[i])
.append(space)
.append(remainder-- > 0 ? " " : "");
return padResult(result.toString().trim(), maxWidth);
}
private int wordsLength(int left, int right, String[] words) {
int wordsLength = 0;
for (int i = left; i <= right; i++) wordsLength += words[i].length();
return wordsLength;
}
private String padResult(String result, int maxWidth) {
return result + blank(maxWidth - result.length());
}
private String blank(int length) {
return new String(new char[length]).replace('\0', ' ');
}
Python 3 Text Justification LeetCode Solution
def fullJustify(self, words, maxWidth):
res, cur, num_of_letters = [], [], 0
for w in words:
if num_of_letters + len(w) + len(cur) > maxWidth:
for i in range(maxWidth - num_of_letters):
cur[i%(len(cur)-1 or 1)] += ' '
res.append(''.join(cur))
cur, num_of_letters = [], 0
cur += [w]
num_of_letters += len(w)
return res + [' '.join(cur).ljust(maxWidth)]
Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1
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