Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Word Wrap Problem
A sequence of words is given, there is a limit on the number of characters for each line. By putting line breaks, in such a way that lines are printed clearly.
The lines must be balanced, when some lines have lots of extra spaces and some lines are containing a small number of extra spaces, it will balance them to separate lines. It tries to use the same number of extra spaces to make them balanced.
This algorithm will produce how many words can be placed in one line, and how many lines are needed.
Input and Output
Input:
The length of words for each line. {3, 2, 2, 5}. The max width is 6.
Output:
Line number 1: Word Number: 1 to 1 (only one word)
Line number 2: Word Number: 2 to 3 (Second and 3rd word)
Line number 3: Word Number: 4 to 4 (4th word)Algorithm
wordWrap(wordLenArr, size, maxWidth)
Input − The word length array, size of the array and the maximum width of the word.
Output − List of how many words will place per line.
Begin define two square matrix extraSpace and lineCost of order (size + 1) define two array totalCost and solution of size (size + 1) for i := 1 to size, do extraSpace[i, i] := maxWidth – wordLenArr[i - 1] for j := i+1 to size, do extraSpace[i, j] := extraSpace[i, j-1] – wordLenArr[j - 1] - 1 done done for i := 1 to size, do for j := i+1 to size, do if extraSpace[i, j] < 0, then lineCost[i, j] = ∞ else if j = size and extraSpace[i, j] >= 0, then lineCost[i, j] := 0 else linCost[i, j] := extraSpace[i, j]^2 done done totalCost[0] := 0 for j := 1 to size, do totalCost[j] := ∞ for i := 1 to j, do if totalCost[i-1] ≠∞ and linCost[i, j] ≠ ∞ and (totalCost[i-1] + lineCost[i,j] < totalCost[j]), then totalCost[i – 1] := totalCost[i – 1] + lineCost[i, j] solution[j] := i done done display the solution matrix End
Example
#include<iostream>
using namespace std;
int dispSolution (int solution[], int size) {
int k;
if (solution[size] == 1)
k = 1;
else
k = dispSolution (solution, solution[size]-1) + 1;
cout << "Line number "<< k << ": Word Number: " <<solution[size]<<" to "<< size << endl;
return k;
}
void wordWrap(int wordLenArr[], int size, int maxWidth) {
int extraSpace[size+1][size+1];
int lineCost[size+1][size+1];
int totalCost[size+1];
int solution[size+1];
for(int i = 1; i<=size; i++) { //find extra space for all lines
extraSpace[i][i] = maxWidth - wordLenArr[i-1];
for(int j = i+1; j<=size; j++) { //extra space when word i to j are in single line
extraSpace[i][j] = extraSpace[i][j-1] - wordLenArr[j-1] - 1;
}
}
for (int i = 1; i <= size; i++) { //find line cost for previously created extra spaces array
for (int j = i; j <= size; j++) {
if (extraSpace[i][j] < 0)
lineCost[i][j] = INT_MAX;
else if (j == size && extraSpace[i][j] >= 0)
lineCost[i][j] = 0;
else
lineCost[i][j] = extraSpace[i][j]*extraSpace[i][j];
}
}
totalCost[0] = 0;
for (int j = 1; j <= size; j++) { //find minimum cost for words
totalCost[j] = INT_MAX;
for (int i = 1; i <= j; i++) {
if (totalCost[i-1] != INT_MAX && lineCost[i][j] != INT_MAX && (totalCost[i-1] + lineCost[i][j] < totalCost[j])){
totalCost[j] = totalCost[i-1] + lineCost[i][j];
solution[j] = i;
}
}
}
dispSolution(solution, size);
}
main() {
int wordLenArr[] = {3, 2, 2, 5};
int n = 4;
int maxWidth = 6;
wordWrap (wordLenArr, n, maxWidth);
}Output
Line number 1: Word Number: 1 to 1 Line number 2: Word Number: 2 to 3 Line number 3: Word Number: 4 to 4
Updated on: 17-Jun-2020
1K+ Views
Advertisements