Justify the given Text based on the given width of each line
Given a string str and width of each line as L, the task is to justify the string such that each line of justified text is of length L with help of space (‘ ‘) and the last line should be left-justified.
Examples:
Input: str = “GeeksforGeek is the best computer science portal for geeks.”, L = 16
Output:
GeeksforGeek__is
the_________best
computer_science
portal_______for
geeks.__________Input: str = “The quick brown fox jumps over the lazy dog.”, L = 11
Output:
The___quick
brown___fox
jumped_over
the____lazy
dogs.______
The symbol _ denotes a space.
Approach:
The number of spaces between words in each line cannot be computed until a complete set of words in that line is known. We will solve this problem on a line-by-line basis.
- Split the given text into words
- Firstly select the words which can be inserted in each line including a space between each word. For that, the sum of length of included words with one space between them must be smaller than or equal to L.
- Now count the number of spaces needed to make the length of each line L and distribute the spaces evenly.
- Repeat above steps for next set of words.
- For the last line spaces must be assigned at the end as the last line must be left-justified.
Below is the implementation of the above approach:
// C++ program to Justify the given Text// according to the given width of each line #include "bits/stdc++.h"using namespace std; // Function to join the words// with spaces spread evenlystring JoinALineWithSpace( vector<string>& words, int start, int end, int num_spaces){ // Number of words in current line int num_words_curr_line = end - start + 1; // String to store the justified text string line; for (int i = start; i < end; i++) { line += words[i]; --num_words_curr_line; // Count number of current space needed int num_curr_space = ceil((double)(num_spaces) / num_words_curr_line); // Insert spaces in string line line.append(num_curr_space, ' '); // Delete the spaces inserted in line num_spaces -= num_curr_space; } // Insert word to string line line += words[end]; line.append(num_spaces, ' '); // Return justified text return line;} // Function that justify the words of// sentence of length of line Lvector<string> JustifyText( vector<string>& words, int L){ int curr_line_start = 0; int num_words_curr_line = 0; int curr_line_length = 0; // To store the justified text vector<string> result; // Traversing the words array for (int i = 0; i < words.size(); i++) { // curr_line_start is the first word // in the current line, and i is // used to identify the last word ++num_words_curr_line; int lookahead_line_length = curr_line_length + words[i].size() + (num_words_curr_line - 1); // If by including the words length becomes L, // then that set of words is justified // and add the justified text to result if (lookahead_line_length == L) { // Justify the set of words string ans = JoinALineWithSpace( words, curr_line_start, i, i - curr_line_start); // Store the justified text in result result.emplace_back(ans); // Start the current line // with next index curr_line_start = i + 1; // num of words in the current line // and current line length set to 0 num_words_curr_line = 0; curr_line_length = 0; } // If by including the words such that // length of words becomes greater than L, // then hat set is justified with // one less word and add the // justified text to result else if (lookahead_line_length > L) { // Justify the set of words string ans = JoinALineWithSpace( words, curr_line_start, i - 1, L - curr_line_length); // Store the justified text in result result.emplace_back(ans); // Current line set to current word curr_line_start = i; // Number of words set to 1 num_words_curr_line = 1; // Current line length set // to current word length curr_line_length = words[i].size(); } // If length is less than L then, // add the word to current line length else { curr_line_length += words[i].size(); } } // Last line is to be left-aligned if (num_words_curr_line > 0) { string line = JoinALineWithSpace( words, curr_line_start, words.size() - 1, num_words_curr_line - 1); line.append( L - curr_line_length - (num_words_curr_line - 1), ' '); // Insert the last line // left-aligned to result result.emplace_back(line); } // Return result return result;} // Function to insert words// of sentencevector<string> splitWords(string str){ vector<string> words; string a = ""; for (int i = 0; str[i]; i++) { // Add char to string a // to get the word if (str[i] != ' ') { a += str[i]; } // If a space occurs // split the words and // add it to vector else { words.push_back(a); a = ""; } } // Push_back the last word // of sentence words.push_back(a); // Return the vector of // words extracted from // string return words;} // Function to print justified textvoid printJustifiedText(vector<string>& result){ for (auto& it : result) { cout << it << endl; }} // Function to call the justificationvoid justifyTheText(string str, int L){ vector<string> words; // Inserting words from // given string words = splitWords(str); // Function call to // justify the text vector<string> result = JustifyText(words, L); // Print the justified // text printJustifiedText(result);} // Driver Codeint main(){ string str = "GeeksforGeek is the best" " computer science portal" " for geeks."; int L = 16; justifyTheText(str, L); return 0;} |
Output:
GeeksforGeek is the best computer science portal for geeks.
Time Complexity: O(N), where N = length of string
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