Added Word Wrap Problem

Algorithm/DP/readme.md

@@ -43,4 +43,7 @@ be avoided by constructinng a temporary array dp[] and memoizing the computed va
 
 - Minimum Number of Deletions To Make a String Palindrome ---->[C++](/Code/C++/min_deletions_to_make_string_palindrome.cpp)
 
-- Climbing Stairs ----> [C++](/Code/C++/Climbing_Stairs.cpp)
+- Climbing Stairs ----> [C++](/Code/C++/Climbing_Stairs.cpp)
+
+- Word Wrap Problem ----> [C++](/Code/C++/word_wrap.cpp)
+

Code/C++/word_wrap.cpp

@@ -0,0 +1,136 @@
+/*
+-WORD WRAP PROBLEM
+-Given an array of size n that contains size of words and maximum width of a line. Arrange the words in such a way that each line does not exceeds the maximum width.
+-Give space after each word except the last one(you can give extra spaces as well)
+-Print 2*L space separated integers, L being the number of lines required to adjust the words with maximum width that shows from which word to word in each line
+
+-DYNAMIC APPROACH
+-Make all possible lines and store cost of each line in a two dimensional table
+-Optimize the final cost,i.e,calculate the minimum cost arrangement
+-Keep track of from where the optimized cost is coming from 
+-Print the optimized cost recursively 
+*/
+
+#include <bits/stdc++.h>
+#define inf INT_MAX
+#define ll long long int 
+using namespace std;
+//recursive function to print the required soluton,  uses p[] to print the solution.
+ll printwordwrap(ll p[],ll n){
+        //starts traversing from back till it get 1
+	//once 1 is achieved it prints the required value and from where it came from
+	ll lineno;
+	if(p[n]==1){
+		lineno=1;
+		cout<<"Line number "<<lineno<<": From word number "<<p[n]<<" to "<<n<<"\n";
+	}
+	else{
+		lineno=printwordwrap(p,p[n]-1)+1;
+	        cout<<"Line number "<<lineno<<": From word number "<<p[n]<<" to "<<n<<"\n";
+	}
+	return lineno;
+}
+
+int main() {
+	//no. of words
+	ll n;
+	cin>>n;
+	//length of each word
+	ll a[n+1];
+	for(ll i=1;i<=n;i++)
+	cin>>a[i];
+        //max width of a line
+	ll max;
+	cin>>max;
+	//space[.][.] contains the number of spaces in a single line
+	ll space[n+1][n+1];
+	//cost[.][.] contains the cost of particular line
+	ll cost[n+1][n+1];
+	//opcost[.] contains the cost of optimized solution
+	ll opcost[n+1];
+	//p[.] for printing the final solution through recursive function printp(...)
+	//it keeps track of the path,i.e, from where the optimized cost is coming
+	ll p[n+1];
+	//loop for calculating space[.][.] when words are arranged in a line
+	for(ll i=1;i<=n;i++)
+	{
+		space[i][i]=max-a[i];
+		for(ll j=i+1;j<=n;j++)
+		{
+			space[i][j]=space[i][j-1]-a[j]-1;
+		}
+	}
+	//loop for calculating cost of particular line when words are arranged in asingle line
+	for(ll i=1;i<=n;i++)
+	{
+		for(ll j=i;j<=n;j++)
+		{
+			if(space[i][j]<0)
+			{
+			//to avoid the condition when words cannot be arranged in one line
+				cost[i][j]=inf;
+			}
+			else if(j==n && space[i][j]>=0)
+			cost[i][j]=0;
+			else cost[i][j]=space[i][j]*space[i][j];
+		}
+	}
+
+	opcost[0]=0;
+	//loop for calculating optimized cost and arrangement of words
+	for(ll i=1;i<=n;i++)
+	{
+		opcost[i]=inf;
+		for(ll j=1;j<=i;j++)
+		{
+			if(opcost[j-1]!=inf && cost[j][i]!=inf && opcost[j-1]+cost[j][i] < opcost[i])
+			{
+				opcost[i]=opcost[j-1]+cost[j][i];
+				p[i]=j;
+			}
+		}
+	}
+	//calling printwordwrap(...)
+	printwordwrap(p,n);
+	cout<<"\n";
+	return 0;
+}
+
+
+/*
+
+TEST CASES
+
+-Test-Case 1
+
+Input:
+18
+7 2 4 2 10 4 6 2 7 2 1 9 3 2 10 4 2 2
+20
+
+Output:
+Line number 1: From word number 1 to 3
+Line number 2: From word number 4 to 6
+Line number 3: From word number 7 to 10
+Line number 4: From word number 11 to 14
+Line number 5: From word number 15 to 17
+Line number 6: From word number 18 to 18
+
+
+-Test-Case 2
+
+Input:
+6
+4 2 2 7 2 4 
+16
+
+Output:
+Line number 1: From word number 1 to 3
+Line number 2: From word number 4 to 6
+
+
+Time Complexity - O(n*n)
+Space Complexity - O(n*n)
+
+*/
+