new question added

Author: arorarahul

README.md

@@ -378,6 +378,7 @@ linear array, array of length 1 less than total can be taken and a pattern can b
 - [Find the longest increasing subsequence](/dynamic-programming/question13.c)
 - [Find the longest decreasing subsequence](/dynamic-programming/question14.c)
 - [Perfect hill longest subsequence](/dynamic-programming/question15.c)
+- [Given two words, word1 and word2, find min operations to convert word1 to word2 with some given set of rules](/dynamic-programming/question16.c)
 
 
 ## Some important concepts to solve algos better

dynamic-programming/a.exe

@@ -0,0 +1,94 @@
+/*
+Given two words, word1 and word2, 
+find min operations to convert word1 to word2 with some given set of rules:
+
+1) insert a character
+2) delete a character
+3) replace a character
+
+METHOD1:
+Naive approach: This approach is to check each and every possibility on a character. EG:
+Str1: data
+str2: dent
+We will compare the first two characters and if they are equal, if they are equal, then we
+simply move on to the next character, if they are not equal we will consider all the possibilities
+of deleting, replacing and inserting the element and compute the result recursively. In the end 
+we check what method is taking minimum number of operations.
+
+Time complexity: O(3^n) //n being the number of elements and there are 3 kind of operations that
+can be applied for each element
+Space complexity: O(1)
+
+METHOD2:
+Dynamic Programming: Here in the above method if a tree is made for each possibility of replacing,
+inserting and deleting as branches we will get repeated sub problems, so in a way we can save them
+in a table and rather solve them in O(1) time.
+
+Therefore we simply create a mXn matrix 
+
+c(i,j) = 	c(i-1,j-1) if match,
+				if no match  
+			min{
+				1 + c(i-1,j-1) if replaced, 
+				1 + c(i-1,j) if delete,  
+				1 + c(i,j-1) if insert    
+		 	}
+
+Therefore, we make a matrix of size m + 1 X n+1 in case length of two strings is m X n.
+Now, imagine matrix like this
+	0	M 	A 	R 	C 	H
+0   0	1	2	3	4	5    //the first row and column means convert C to 0 converting CA to 0 and so on
+C 	1
+A 	2
+R 	3
+T 	4
+
+If there is a match, we simply pick the diagnal element now when traversing, if there is no match,
+we take the min of top left and diagnal and add 1 to it. Its logical as if there is a match, the
+size of the problem is reduced by 1 on both sides as per recursive eqn.
+
+If there is no match, we have to compute all problems as per recursive equation which has already been
+computed in the matrix
+
+Time complexity: O(mn)
+Space complexity: O(mn)
+*/
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+int findMin(int a, int b, int c){
+	return (a < b)?((a<c)?a:c):((b<c)?b:c);
+}
+
+int findEditDistance(char *str1, char *str2){
+	int len1 = strlen(str1);
+	int len2 = strlen(str2);
+	printf("length of two strings are %d and %d\n", len1, len2);
+	int editDis[len1 + 1][len2 + 1];
+	int i,j;
+	for(i=0; i<=len1;i++){
+		for(j=0;j<=len2;j++){
+			if(i == 0){
+				editDis[i][j] = j; 
+			}else if(j == 0){
+				editDis[i][j] = i;
+			}else if(str1[i-1] == str2[j-1]){
+				editDis[i][j] = editDis[i-1][j-1];
+			}else{
+				editDis[i][j] = 1 + findMin(editDis[i-1][j],editDis[i][j-1],editDis[i-1][j-1]);
+			}
+		}
+	}
+	return editDis[len1][len2];
+}
+
+int main(){
+	char str1[] = "CART";
+	char str2[] = "MARCH";
+
+	int operations = findEditDistance(str1, str2);
+	printf("min number of operations req are: %d\n", operations);
+
+	return 0;
+}