diff --git a/Dynamic Programming/Edit_Distance.java b/Dynamic Programming/Edit_Distance.java
index 00835bc5c..87c504218 100644
--- a/Dynamic Programming/Edit_Distance.java	
+++ b/Dynamic Programming/Edit_Distance.java	
@@ -1,26 +1,41 @@
     /**
     Author : SUBHAM SANGHAI
     A Dynamic Programming based solution for Edit Distance problem In Java
-    Edit distance is a way of quantifying how dissimilar two strings (e.g., words) are to one another, 
-    by counting the minimum number of operations required to transform one string into the other
     **/
-    import java.util.HashMap;
-    import java.util.Map;
+
+    /**Description of Edit Distance with an Example:
+
+    Edit distance is a way of quantifying how dissimilar two strings (e.g., words) are to one another, 
+    by counting the minimum number of operations required to transform one string into the other. The
+    distance operations are the removal, insertion, or substitution of a character in the string.
+
+
+    The Distance between "kitten" and "sitting" is 3. A minimal edit script that transforms the former into the latter is:
+
+    kitten → sitten (substitution of "s" for "k")
+    sitten → sittin (substitution of "i" for "e")
+    sittin → sitting (insertion of "g" at the end).**/
+
     import java.util.Scanner;
     public class Edit_Distance
     {
+
+      
+
       public static int minDistance(String word1, String word2) 
       {
     	int len1 = word1.length();
     	int len2 = word2.length();
      	// len1+1, len2+1, because finally return dp[len1][len2]
     	int[][] dp = new int[len1 + 1][len2 + 1];
-     
+     	/* If second string is empty, the only option is to
+   	  insert all characters of first string into second*/
     	for (int i = 0; i <= len1; i++) 
     	{
     		dp[i][0] = i;
     	}
-     
+     	/* If first string is empty, the only option is to
+   	  insert all characters of second string into first*/
     	for (int j = 0; j <= len2; j++) 
     	{
     		dp[0][j] = j;
@@ -40,6 +55,8 @@
     			} 
     			else 
     			{
+			/* if two characters are different ,
+			then take the minimum of the various operations(i.e insertion,removal,substitution)*/
     				int replace = dp[i][j] + 1;
     				int insert = dp[i][j + 1] + 1;
     				int delete = dp[i + 1][j] + 1;
@@ -50,8 +67,11 @@
     			}
     		}
     	}
+	/* return the final answer , after traversing through both the strings*/
     	return dp[len1][len2];
       }
+
+
     	// Driver program to test above function
     	public static void main(String args[])
     	{