Word Wrap Problem | DP-19

Given a sequence of words, and a limit on the number of characters that can be put in one line (line width). Put line breaks in the given sequence such that the lines are printed neatly. Assume that the length of each word is smaller than the line width.
Word processors like MS Word do the task of placing line breaks. The idea is to have balanced lines. In other words, do not have a few lines with lots of extra space and some lines with a small amount of extra space. 
 

The extra spaces includes spaces put at the end of every line except the last one.  
The problem is to minimize the following total cost.
 Cost of a line = (Number of extra spaces in the line)^3
 Total Cost = Sum of costs for all lines

For example, consider the following string and line width M = 15
 "Geeks for Geeks presents word wrap problem" 
     
Following is the optimized arrangement of words in 3 lines
Geeks for Geeks
presents word
wrap problem 

The total extra spaces in line 1, line 2 and line 3 are 0, 2 and 3 respectively. 
So optimal value of total cost is 0 + 2*2*2 + 3*3*3 = 35

Please note that the total cost function is not sum of extra spaces, but sum of cubes (or square is also used) of extra spaces. The idea behind this cost function is to balance the spaces among lines. For example, consider the following two arrangement of same set of words:
1) There are 3 lines. One line has 3 extra spaces and all other lines have 0 extra spaces. Total extra spaces = 3 + 0 + 0 = 3. Total cost = 3*3*3 + 0*0*0 + 0*0*0 = 27.
2) There are 3 lines. Each of the 3 lines has one extra space. Total extra spaces = 1 + 1 + 1 = 3. Total cost = 1*1*1 + 1*1*1 + 1*1*1 = 3.
Total extra spaces are 3 in both scenarios, but second arrangement should be preferred because extra spaces are balanced in all three lines. The cost function with cubic sum serves the purpose because the value of total cost in second scenario is less.
 

Method 1 (Greedy Solution) 
The greedy solution is to place as many words as possible in the first line. Then do the same thing for the second line and so on until all words are placed. This solution gives optimal solution for many cases, but doesn’t give optimal solution in all cases. For example, consider the following string “aaa bb cc ddddd” and line width as 6. Greedy method will produce following output. 

aaa bb 
cc 
ddddd

Extra spaces in the above 3 lines are 0, 4 and 1 respectively. So total cost is 0 + 64 + 1 = 65.
But the above solution is not the best solution. Following arrangement has more balanced spaces. Therefore less value of total cost function.

aaa
bb cc
ddddd

Extra spaces in the above 3 lines are 3, 1 and 1 respectively. So total cost is 27 + 1 + 1 = 29.
Despite being sub-optimal in some cases, the greedy approach is used by many word processors like MS Word and OpenOffice.org Writer.

Method 2 (Recursive Approach with memoization)

The problem can be solved using a divide and conquer (recursive) approach. The algorithm for the same is mentioned below:

1. We recur for each word starting with first word, and remaining length of the line (initially k).
2. The last word would be the base case:
    We check if we can put it on same line:
        if yes, then we return cost as 0.
        if no, we return cost of current line based on its remaining length.
3. For non-last words, we have to check if it can fit in the current line:
    if yes, then we have two choices i.e. whether to put it in same line or next line.
        if we put it on next line: cost1 = square(remLength) + cost of putting word on next line.
        if we put it on same line: cost2 = cost of putting word on same line.
        return min(cost1, cost2)
    if no, then we have to put it on next line:
        return cost of putting word on next line
4. We use memoization table of size n (number of words) * k (line length), to keep track of already visited positions.

10

This program solves the word wrap problem using dynamic programming and memoization. The word wrap problem involves formatting a given sequence of words into lines such that each line has a maximum width limit (k). The goal is to minimize the total number of extra spaces at the end of each line.

The program defines two functions:

solveWordWrap: This is the main function that takes in an array of words, the number of words, and the maximum line width as input. It initializes a memoization table and then calls the solveWordWrapUsingMemo function to compute the minimum extra spaces needed.
solveWordWrapUsingMemo: This function takes in the same inputs as solveWordWrap but also includes a memoization table to store the results of subproblems that have already been solved. It uses recursion to break down the problem into subproblems, each with one less word to consider. The memoization table is used to avoid recomputing the solution to subproblems that have already been solved.

Time Complexity: The time complexity of the program is O(nk), where n is the number of words and k is the maximum line width. This is because the program computes the solution to each subproblem once and stores the result in the memoization table, so each subproblem is only solved once. Since there are nk subproblems, the overall time complexity is O(n*k).

Space Complexity: The space complexity of the program is also O(n*k), because the memoization table is a 2D array with n rows and k+1 columns.

Method 3 (Dynamic Programming) 
The following Dynamic approach strictly follows the algorithm given in solution of Cormen book. First we compute costs of all possible lines in a 2D table lc[][]. The value lc[i][j] indicates the cost to put words from i to j in a single line where i and j are indexes of words in the input sequences. If a sequence of words from i to j cannot fit in a single line, then lc[i][j] is considered infinite (to avoid it from being a part of the solution). Once we have the lc[][] table constructed, we can calculate total cost using following recursive formula. In the following formula, C[j] is the optimized total cost for arranging words from 1 to j. 
 

The above recursion has overlapping subproblem property. For example, the solution of subproblem c(2) is used by c(3), C(4) and so on. So Dynamic Programming is used to store the results of subproblems. The array c[] can be computed from left to right, since each value depends only on earlier values. 
To print the output, we keep track of what words go on what lines, we can keep a parallel p array that points to where each c value came from. The last line starts at word p[n] and goes through word n. The previous line starts at word p[p[n]] and goes through word p[n] – 1, etc. The function printSolution() uses p[] to print the solution. 
In the below program, input is an array l[] that represents lengths of words in a sequence. The value l[i] indicates length of the ith word (i starts from 1) in theinput sequence. 
 

Line number 1: From word no. 1 to 1
Line number 2: From word no. 2 to 3
Line number 3: From word no. 4 to 4

Time Complexity: O(n^2) 
Auxiliary Space: O(n^2) The auxiliary space used in the above program can be optimized to O(n) (See the reference 2 for details)

Word Wrap problem ( Space optimized solution )
References: 
http://en.wikipedia.org/wiki/Word_wrap
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