Word Wrap Problem | DP-19

12 May, 2022
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.The word processors like MS Word do task of placing line breaks. The idea is to have balanced lines. In other words, not have few lines with lots of extra spaces and some lines with small amount of extra spaces.  
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.
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Python3
C#
Javascript
#include <bits/stdc++.h>using namespace std; int solveWordWrapUsingMemo(int words[], int n, int length,                           int wordIndex, int remLength,                           vector<vector<int> > memo); int square(int n) { return n * n; } int solveWordWrapUtil(int words[], int n, int length,                      int wordIndex, int remLength,                      vector<vector<int> > memo){     // base case for last word    if (wordIndex == n - 1) {        memo[wordIndex][remLength]            = words[wordIndex] < remLength                  ? 0                  : square(remLength);        return memo[wordIndex][remLength];    }     int currWord = words[wordIndex];    // if word can fit in the remaining line    if (currWord < remLength) {        return min(solveWordWrapUsingMemo(                       words, n, length, wordIndex + 1,                       remLength == length                           ? remLength - currWord                           : remLength - currWord - 1,                       memo),                    square(remLength)                       + solveWordWrapUsingMemo(                           words, n, length, wordIndex + 1,                           length - currWord, memo));    }    else {        // if word is kept on next line        return square(remLength)               + solveWordWrapUsingMemo(                   words, n, length, wordIndex + 1,                   length - currWord, memo);    }} int solveWordWrapUsingMemo(int words[], int n, int length,                           int wordIndex, int remLength,                           vector<vector<int> > memo){    if (memo[wordIndex][remLength] != -1) {        return memo[wordIndex][remLength];    }     memo[wordIndex][remLength] = solveWordWrapUtil(        words, n, length, wordIndex, remLength, memo);    return memo[wordIndex][remLength];} int solveWordWrap(int words[], int n, int k){     vector<vector<int> > memo(n, vector<int>(k + 1, -1));     return solveWordWrapUsingMemo(words, n, k, 0, k, memo);}int main(){    int words[] = { 3, 2, 2, 5 };    int n = sizeof(words) / sizeof(words[0]);    int k = 6;     cout << solveWordWrap(words, n, k);    return 0;}/* This Code is contributed by Tapesh (tapeshdua420) */
/*package whatever //do not write package name here */ import java.io.*;import java.util.Arrays; public class WordWrapDpMemo {     private int solveWordWrap(int[] nums, int k) {        int[][] memo = new int[nums.length][k + 1];        for (int i = 0; i < nums.length; i++) {            Arrays.fill(memo[i], -1);        }        return solveWordWrapUsingMemo(nums, nums.length, k, 0, k, memo);    }     private int solveWordWrap(int[] words, int n, int length, int wordIndex, int remLength, int[][] memo) {         //base case for last word        if (wordIndex == n - 1) {            memo[wordIndex][remLength] = words[wordIndex] < remLength ? 0 : square(remLength);            return memo[wordIndex][remLength];        }         int currWord = words[wordIndex];        //if word can fit in the remaining line        if (currWord < remLength) {            return Math.min(                    //if word is kept on same line                    solveWordWrapUsingMemo(words, n, length, wordIndex + 1, remLength == length ? remLength - currWord : remLength - currWord - 1, memo),                    //if word is kept on next line                    square(remLength) + solveWordWrapUsingMemo(words, n, length, wordIndex + 1, length - currWord, memo)            );        } else {            //if word is kept on next line            return square(remLength) + solveWordWrapUsingMemo(words, n, length, wordIndex + 1, length - currWord, memo);        }    }     private int solveWordWrapUsingMemo(int[] words, int n, int length, int wordIndex, int remLength, int[][] memo) {        if (memo[wordIndex][remLength] != -1) {            return memo[wordIndex][remLength];        }         memo[wordIndex][remLength] = solveWordWrap(words, n, length, wordIndex, remLength, memo);        return memo[wordIndex][remLength];    }     private int square(int n) {        return n * n;    }     public static void main(String[] args) {        System.out.println(new WordWrapDpMemo().solveWordWrap(new int[]{3, 2, 2, 5}, 6));    }}
# Python program for the above approachdef square(n) :    return n * n  def solveWordWrapUtil(words, n, length,                      wordIndex, remLength, memo) :      # base case for last word    if (wordIndex == n - 1) :        memo[wordIndex][remLength] = 0 if (words[wordIndex] < remLength) else square(remLength)        return memo[wordIndex][remLength]           currWord = words[wordIndex]    # if word can fit in the remaining line    if (currWord < remLength) :        return min(solveWordWrapUsingMemo(                       words, n, length, wordIndex + 1,                       remLength - currWord if (remLength == length) else remLength - currWord - 1,                       memo),                     square(remLength)                       + solveWordWrapUsingMemo(                           words, n, length, wordIndex + 1,                           length - currWord, memo))         else :        # if word is kept on next line        return (square(remLength)               + solveWordWrapUsingMemo(                   words, n, length, wordIndex + 1,                   length - currWord, memo))     def solveWordWrapUsingMemo(words, n, length,                           wordIndex, remLength, memo) :                                    if (memo[wordIndex][remLength] != -1) :        return memo[wordIndex][remLength]         memo[wordIndex][remLength] = (solveWordWrapUtil(        words, n, length, wordIndex, remLength, memo))    return memo[wordIndex][remLength] def solveWordWrap(words, n, k) :    memo = [[10]* (k + 1)]* n    return solveWordWrapUsingMemo(words, n, k, 0, k, memo) # Driver Codeif __name__ == "__main__":     words = [ 3, 2, 2, 5 ]    n = len(words)    k = 6      print(solveWordWrap(words, n, k))     # This code is contributed by sanjoy_62.
/*package whatever //do not write package name here */using System;using System.Collections.Generic; public class WordWrapDpMemo {     private int solveWordWrap(int[] nums, int k)    {        int[,] memo = new int[nums.Length ,k + 1];        for (int i = 0; i < memo.GetLength(0); i++)        {            for(int j = 0; j < memo.GetLength(1); j++)            memo[i, j] = -1;        }        return solveWordWrapUsingMemo(nums, nums.Length,                                      k, 0, k, memo);    }     private int solveWordWrap(int[] words, int n,                              int length, int wordIndex,                              int remLength, int[,] memo)    {         // base case for last word        if (wordIndex == n - 1) {            memo[wordIndex, remLength] = words[wordIndex] <              remLength ? 0 : square(remLength);            return memo[wordIndex, remLength];        }         int currWord = words[wordIndex];               // if word can fit in the remaining line        if (currWord < remLength) {            return Math.Min(                                   // if word is kept on same line                    solveWordWrapUsingMemo(words, n, length, wordIndex + 1,                            remLength == length ? remLength -                                           currWord : remLength -                                           currWord - 1, memo),                                   // if word is kept on next line                    square(remLength)                            + solveWordWrapUsingMemo(words, n,                                                     length,                                                     wordIndex + 1,                                                     length - currWord,                                                     memo));        } else {            // if word is kept on next line            return square(remLength) +              solveWordWrapUsingMemo(words, n, length,                                     wordIndex + 1,                                     length - currWord, memo);        }    }     private int solveWordWrapUsingMemo(int[] words, int n,                                       int length, int wordIndex,                                       int remLength, int[,] memo)    {        if (memo[wordIndex,remLength] != -1)        {            return memo[wordIndex,remLength];        }         memo[wordIndex,remLength] = solveWordWrap(words,                                                  n, length,                                                  wordIndex,                                                  remLength, memo);        return memo[wordIndex, remLength];    }     private int square(int n) {        return n * n;    }     public static void Main(String[] args) {        Console.WriteLine(new WordWrapDpMemo().                          solveWordWrap(new int[] { 3, 2, 2, 5 }, 6));    }} // This code is contributed by gauravrajput1
<script>/*package whatever //do not write package name here */function solveWordWrap(nums , k) {        var memo = Array(nums.length).fill().map(()=>Array(k + 1).fill(-1));        return solveWordWrapUsingMemo(nums, nums.length, k, 0, k, memo);    }     function solveWordWrap1(words , n , length , wordIndex , remLength,  memo) {         // base case for last word        if (wordIndex == n - 1) {            memo[wordIndex][remLength] = words[wordIndex] < remLength ? 0 : square(remLength);            return memo[wordIndex][remLength];        }         var currWord = words[wordIndex];        // if word can fit in the remaining line        if (currWord < remLength) {            return Math.min(                    // if word is kept on same line                    solveWordWrapUsingMemo(words, n, length, wordIndex + 1,                            remLength == length ? remLength - currWord : remLength - currWord - 1, memo),                    // if word is kept on next line                    square(remLength)                            + solveWordWrapUsingMemo(words, n, length, wordIndex + 1, length - currWord, memo));        } else {            // if word is kept on next line            return square(remLength) + solveWordWrapUsingMemo(words, n, length, wordIndex + 1, length - currWord, memo);        }    }     function solveWordWrapUsingMemo(words , n , length , wordIndex , remLength,  memo) {        //if (memo[wordIndex][remLength] != -1) {        //    return memo[wordIndex][remLength];        //}         memo[wordIndex][remLength] = solveWordWrap1(words, n, length, wordIndex, remLength, memo);        return memo[wordIndex][remLength];    }     function square(n) {        return n * n;    }     var arr = [ 3, 2, 2, 5 ];        document.write(solveWordWrap(arr, 6)); // This code is contributed by gauravrajput1</script>
10
 
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.  
C++
C
Java
Python3
C#
Javascript
// A Dynamic programming solution for Word Wrap Problem#include <bits/stdc++.h>using namespace std;#define INF INT_MAX // A utility function to print the solutionint printSolution (int p[], int n); // l[] represents lengths of different words in input sequence.// For example, l[] = {3, 2, 2, 5} is for a sentence like// "aaa bb cc ddddd". n is size of l[] and M is line width// (maximum no. of characters that can fit in a line)void solveWordWrap (int l[], int n, int M){    // For simplicity, 1 extra space is used in all below arrays     // extras[i][j] will have number of extra spaces if words from i    // to j are put in a single line    int extras[n+1][n+1];     // lc[i][j] will have cost of a line which has words from    // i to j    int lc[n+1][n+1];     // c[i] will have total cost of optimal arrangement of words    // from 1 to i    int c[n+1];     // p[] is used to print the solution.    int p[n+1];     int i, j;     // calculate extra spaces in a single line. The value extra[i][j]    // indicates extra spaces if words from word number i to j are    // placed in a single line    for (i = 1; i <= n; i++)    {        extras[i][i] = M - l[i-1];        for (j = i+1; j <= n; j++)            extras[i][j] = extras[i][j-1] - l[j-1] - 1;    }     // Calculate line cost corresponding to the above calculated extra    // spaces. The value lc[i][j] indicates cost of putting words from    // word number i to j in a single line    for (i = 1; i <= n; i++)    {        for (j = i; j <= n; j++)        {            if (extras[i][j] < 0)                lc[i][j] = INF;            else if (j == n && extras[i][j] >= 0)                lc[i][j] = 0;            else                lc[i][j] = extras[i][j]*extras[i][j];        }    }     // Calculate minimum cost and find minimum cost arrangement.    // The value c[j] indicates optimized cost to arrange words    // from word number 1 to j.    c[0] = 0;    for (j = 1; j <= n; j++)    {        c[j] = INF;        for (i = 1; i <= j; i++)        {            if (c[i-1] != INF && lc[i][j] != INF &&                           (c[i-1] + lc[i][j] < c[j]))            {                c[j] = c[i-1] + lc[i][j];                p[j] = i;            }        }    }     printSolution(p, n);} int printSolution (int p[], int n){    int k;    if (p[n] == 1)        k = 1;    else        k = printSolution (p, p[n]-1) + 1;    cout<<"Line number "<<k<<": From word no. "<<p[n]<<" to "<<n<<endl;    return k;} // Driver program to test above functionsint main(){    int l[] = {3, 2, 2, 5};    int n = sizeof(l)/sizeof(l[0]);    int M = 6;    solveWordWrap (l, n, M);    return 0;}  //This is code is contributed by rathbhupendra
// A Dynamic programming solution for Word Wrap Problem#include <limits.h>#include <stdio.h>#define INF INT_MAX // A utility function to print the solutionint printSolution (int p[], int n); // l[] represents lengths of different words in input sequence.// For example, l[] = {3, 2, 2, 5} is for a sentence like// "aaa bb cc ddddd". n is size of l[] and M is line width// (maximum no. of characters that can fit in a line)void solveWordWrap (int l[], int n, int M){    // For simplicity, 1 extra space is used in all below arrays     // extras[i][j] will have number of extra spaces if words from i    // to j are put in a single line    int extras[n+1][n+1];      // lc[i][j] will have cost of a line which has words from    // i to j    int lc[n+1][n+1];      // c[i] will have total cost of optimal arrangement of words    // from 1 to i    int c[n+1];     // p[] is used to print the solution.     int p[n+1];     int i, j;     // calculate extra spaces in a single line.  The value extra[i][j]    // indicates extra spaces if words from word number i to j are    // placed in a single line    for (i = 1; i <= n; i++)    {        extras[i][i] = M - l[i-1];        for (j = i+1; j <= n; j++)            extras[i][j] = extras[i][j-1] - l[j-1] - 1;    }     // Calculate line cost corresponding to the above calculated extra    // spaces. The value lc[i][j] indicates cost of putting words from    // word number i to j in a single line    for (i = 1; i <= n; i++)    {        for (j = i; j <= n; j++)        {            if (extras[i][j] < 0)                lc[i][j] = INF;            else if (j == n && extras[i][j] >= 0)                lc[i][j] = 0;            else                lc[i][j] = extras[i][j]*extras[i][j];        }    }     // Calculate minimum cost and find minimum cost arrangement.    //  The value c[j] indicates optimized cost to arrange words    // from word number 1 to j.    c[0] = 0;    for (j = 1; j <= n; j++)    {        c[j] = INF;        for (i = 1; i <= j; i++)        {            if (c[i-1] != INF && lc[i][j] != INF &&               (c[i-1] + lc[i][j] < c[j]))            {                c[j] = c[i-1] + lc[i][j];                p[j] = i;            }        }    }     printSolution(p, n);} int printSolution (int p[], int n){    int k;    if (p[n] == 1)        k = 1;    else        k = printSolution (p, p[n]-1) + 1;    printf ("Line number %d: From word no. %d to %d \n", k, p[n], n);    return k;} // Driver program to test above functionsint main(){    int l[] = {3, 2, 2, 5};    int n = sizeof(l)/sizeof(l[0]);    int M = 6;    solveWordWrap (l, n, M);    return 0;}
// A Dynamic programming solution for// Word Wrap Problem in Javapublic class WordWrap{     final int MAX = Integer.MAX_VALUE;         // A utility function to print the solution    int printSolution (int p[], int n)    {        int k;        if (p[n] == 1)        k = 1;        else        k = printSolution (p, p[n]-1) + 1;        System.out.println("Line number" + " " + k + ": " +                    "From word no." +" "+ p[n] + " " + "to" + " " + n);        return k;    } // l[] represents lengths of different words in input sequence.// For example, l[] = {3, 2, 2, 5} is for a sentence like// "aaa bb cc ddddd". n is size of l[] and M is line width// (maximum no. of characters that can fit in a line)    void solveWordWrap (int l[], int n, int M)    {        // For simplicity, 1 extra space is used in all below arrays             // extras[i][j] will have number of extra spaces if words from i        // to j are put in a single line        int extras[][] = new int[n+1][n+1];             // lc[i][j] will have cost of a line which has words from        // i to j        int lc[][]= new int[n+1][n+1];             // c[i] will have total cost of optimal arrangement of words        // from 1 to i        int c[] = new int[n+1];             // p[] is used to print the solution.        int p[] =new int[n+1];             // calculate extra spaces in a single line. The value extra[i][j]        // indicates extra spaces if words from word number i to j are        // placed in a single line        for (int i = 1; i <= n; i++)        {            extras[i][i] = M - l[i-1];            for (int j = i+1; j <= n; j++)            extras[i][j] = extras[i][j-1] - l[j-1] - 1;        }                 // Calculate line cost corresponding to the above calculated extra        // spaces. The value lc[i][j] indicates cost of putting words from        // word number i to j in a single line        for (int i = 1; i <= n; i++)        {            for (int j = i; j <= n; j++)            {                if (extras[i][j] < 0)                    lc[i][j] = MAX;                else if (j == n && extras[i][j] >= 0)                    lc[i][j] = 0;                else                    lc[i][j] = extras[i][j]*extras[i][j];            }        }                 // Calculate minimum cost and find minimum cost arrangement.        // The value c[j] indicates optimized cost to arrange words        // from word number 1 to j.        c[0] = 0;        for (int j = 1; j <= n; j++)        {            c[j] = MAX;            for (int i = 1; i <= j; i++)            {                if (c[i-1] != MAX && lc[i][j] != MAX &&                   (c[i-1] + lc[i][j] < c[j]))                {                    c[j] = c[i-1] + lc[i][j];                    p[j] = i;                }            }        }             printSolution(p, n);    }     public static void main(String args[])    {        WordWrap w = new WordWrap();        int l[] = {3, 2, 2, 5};        int n = l.length;        int M = 6;        w.solveWordWrap (l, n, M);    }} // This code is contributed by Saket Kumar
# A Dynamic programming solution# for Word Wrap Problem # A utility function to print# the solution# l[] represents lengths of different# words in input sequence. For example,# l[] = {3, 2, 2, 5} is for a sentence# like "aaa bb cc ddddd". n is size of# l[] and M is line width (maximum no.# of characters that can fit in a line)INF = 2147483647def printSolution(p, n):    k = 0    if p[n] == 1:        k = 1    else:        k = printSolution(p, p[n] - 1) + 1    print('Line number ', k, ': From word no. ',                                 p[n], 'to ', n)    return k def solveWordWrap (l, n, M):         # For simplicity, 1 extra space is    # used in all below arrays     # extras[i][j] will have number    # of extra spaces if words from i    # to j are put in a single line    extras = [[0 for i in range(n + 1)]                 for i in range(n + 1)]                      # lc[i][j] will have cost of a line    # which has words from i to j    lc = [[0 for i in range(n + 1)]             for i in range(n + 1)]                  # c[i] will have total cost of    # optimal arrangement of words    # from 1 to i    c = [0 for i in range(n + 1)]         # p[] is used to print the solution.    p = [0 for i in range(n + 1)]         # calculate extra spaces in a single    # line. The value extra[i][j] indicates    # extra spaces if words from word number    # i to j are placed in a single line    for i in range(n + 1):        extras[i][i] = M - l[i - 1]        for j in range(i + 1, n + 1):            extras[i][j] = (extras[i][j - 1] -                                    l[j - 1] - 1)                                         # Calculate line cost corresponding    # to the above calculated extra    # spaces. The value lc[i][j] indicates    # cost of putting words from word number    # i to j in a single line    for i in range(n + 1):        for j in range(i, n + 1):            if extras[i][j] < 0:                lc[i][j] = INF;            elif j == n and extras[i][j] >= 0:                lc[i][j] = 0            else:                lc[i][j] = (extras[i][j] *                            extras[i][j])     # Calculate minimum cost and find    # minimum cost arrangement. The value    # c[j] indicates optimized cost to    # arrange words from word number 1 to j.    c[0] = 0    for j in range(1, n + 1):        c[j] = INF        for i in range(1, j + 1):            if (c[i - 1] != INF and                lc[i][j] != INF and                ((c[i - 1] + lc[i][j]) < c[j])):                c[j] = c[i-1] + lc[i][j]                p[j] = i    printSolution(p, n)     # Driver Codel = [3, 2, 2, 5]n = len(l)M = 6solveWordWrap(l, n, M) # This code is contributed by sahil shelangia
// A Dynamic programming solution for Word Wrap// Problem in Javausing System; public class GFG {     static int MAX = int.MaxValue;         // A utility function to print the solution    static int printSolution (int []p, int n)    {        int k;                 if (p[n] == 1)            k = 1;        else            k = printSolution (p, p[n]-1) + 1;                     Console.WriteLine("Line number" + " "                + k + ": From word no." + " "                + p[n] + " " + "to" + " " + n);        return k;    }     // l[] represents lengths of different    // words in input sequence. For example,    // l[] = {3, 2, 2, 5} is for a sentence    // like "aaa bb cc ddddd". n is size of    // l[] and M is line width (maximum no.    // of characters that can fit in a line)    static void solveWordWrap (int []l, int n,                                        int M)    {                 // For simplicity, 1 extra space        // is used in all below arrays             // extras[i][j] will have number of        // extra spaces if words from i        // to j are put in a single line        int [,]extras = new int[n+1,n+1];             // lc[i][j] will have cost of a line        // which has words from i to j        int [,]lc = new int[n+1,n+1];             // c[i] will have total cost of        // optimal arrangement of words        // from 1 to i        int []c = new int[n+1];             // p[] is used to print the solution.        int []p = new int[n+1];             // calculate extra spaces in a single        // line. The value extra[i][j] indicates        // extra spaces if words from word number        // i to j are placed in a single line        for (int i = 1; i <= n; i++)        {            extras[i,i] = M - l[i-1];                         for (int j = i+1; j <= n; j++)                extras[i,j] = extras[i,j-1]                                 - l[j-1] - 1;        }                 // Calculate line cost corresponding to        // the above calculated extra spaces. The        // value lc[i][j] indicates cost of        // putting words from word number i to        // j in a single line        for (int i = 1; i <= n; i++)        {            for (int j = i; j <= n; j++)            {                if (extras[i,j] < 0)                    lc[i,j] = MAX;                else if (j == n &&                              extras[i,j] >= 0)                    lc[i,j] = 0;                else                    lc[i,j] = extras[i,j]                                 * extras[i,j];            }        }                 // Calculate minimum cost and find        // minimum cost arrangement. The value        // c[j] indicates optimized cost to        // arrange words from word number        // 1 to j.        c[0] = 0;        for (int j = 1; j <= n; j++)        {            c[j] = MAX;            for (int i = 1; i <= j; i++)            {                if (c[i-1] != MAX && lc[i,j]                    != MAX && (c[i-1] + lc[i,j]                                       < c[j]))                {                    c[j] = c[i-1] + lc[i,j];                    p[j] = i;                }            }        }             printSolution(p, n);    }     // Driver code    public static void Main()    {        int []l = {3, 2, 2, 5};        int n = l.Length;        int M = 6;        solveWordWrap (l, n, M);    }} // This code is contributed by nitin mittal.
<script>    // A Dynamic programming solution for    // Word Wrap Problem in Javascript         let MAX = Number.MAX_VALUE;           // A utility function to print the solution    function printSolution (p, n)    {        let k;        if (p[n] == 1)            k = 1;        else            k = printSolution (p, p[n]-1) + 1;        document.write("Line number" + " " + k + ": " +                    "From word no." +" "+ p[n] + " " + "to" + " " + n + "</br>");        return k;    }       // l[] represents lengths of different words in input sequence.    // For example, l[] = {3, 2, 2, 5} is for a sentence like    // "aaa bb cc ddddd". n is size of l[] and M is line width    // (maximum no. of characters that can fit in a line)    function solveWordWrap (l, n, M)    {        // For simplicity, 1 extra space is used in all below arrays               // extras[i][j] will have number of extra spaces if words from i        // to j are put in a single line        let extras = new Array(n+1);               // lc[i][j] will have cost of a line which has words from        // i to j        let lc = new Array(n+1);                 for(let i = 0; i < n + 1; i++)        {            extras[i] = new Array(n + 1);            lc[i] = new Array(n + 1);            for(let j = 0; j < n + 1; j++)            {                extras[i][j] = 0;                lc[i][j] = 0;            }        }               // c[i] will have total cost of optimal arrangement of words        // from 1 to i        let c = new Array(n+1);               // p[] is used to print the solution.        let p = new Array(n+1);               // calculate extra spaces in a single line. The value extra[i][j]        // indicates extra spaces if words from word number i to j are        // placed in a single line        for (let i = 1; i <= n; i++)        {            extras[i][i] = M - l[i-1];            for (let j = i+1; j <= n; j++)                extras[i][j] = extras[i][j-1] - l[j-1] - 1;        }                   // Calculate line cost corresponding to the above calculated extra        // spaces. The value lc[i][j] indicates cost of putting words from        // word number i to j in a single line        for (let i = 1; i <= n; i++)        {            for (let j = i; j <= n; j++)            {                if (extras[i][j] < 0)                    lc[i][j] = MAX;                else if (j == n && extras[i][j] >= 0)                    lc[i][j] = 0;                else                    lc[i][j] = extras[i][j]*extras[i][j];            }        }                   // Calculate minimum cost and find minimum cost arrangement.        // The value c[j] indicates optimized cost to arrange words        // from word number 1 to j.        c[0] = 0;        for (let j = 1; j <= n; j++)        {            c[j] = MAX;            for (let i = 1; i <= j; i++)            {                if (c[i-1] != MAX && lc[i][j] != MAX &&                   (c[i-1] + lc[i][j] < c[j]))                {                    c[j] = c[i-1] + lc[i][j];                    p[j] = i;                }            }        }               printSolution(p, n);    }         let l = [3, 2, 2, 5];    let n = l.length;    let M = 6;    solveWordWrap (l, n, M);         // This code is contributed by mukesh07.</script>
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 cane be optimized to O(n) (See the reference 2 for details)Word Wrap problem ( Space optimized solution )References: http://en.wikipedia.org/wiki/Word_wrapPlease write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
nitin mittal
sahilshelangia
rathbhupendra
mukesh07
aaditya pandilwar
GauravRajput1
hardikgulati96
tapeshdua420
sanjoy_62
Microsoft
Dynamic Programming
Microsoft
Dynamic Programming
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