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+// 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 solution
+int 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 functions
+int main()
+{
+	int l[] = {3, 2, 2, 5};
+	int n = sizeof(l)/sizeof(l[0]);
+	int M = 6;
+	solveWordWrap (l, n, M);
+	return 0;
+}
+
+