new question added

Author: arorarahul

README.md

@@ -176,7 +176,8 @@ total ways to reach the nth step  f(n) = f(n-1) + f(n-2)
 i.e from n-1 it can take you 1 step and from n-2 it can take you only 1 step of size 2. Therefore, this is
 nothing but fibonacci series.
 - In some algos involving DP you can start from n and in that case answer to n is dependent on n-1 and so on.
-
+- Most problems involving strings can be taken as S(i,j), either we compare last characters then if equal
+it gets converted to i-1,j-1 or we take min or max of i,j-1 and i-1,j
 
 # Topic0: Programming Questions
 
@@ -395,6 +396,8 @@ nothing but fibonacci series.
 - [Partition problem](/dynamic-programming/question20.c)
 - [Find the longest palindromic subsequence](/dynamic-programming/question21.c)
 - [Given n-stairs, how many number of ways a person can climb to top from bottom using 1 step or 2 steps](/dynamic-programming/question22.c)
+- [Longest non-overlapping repeating sub string](/dynamic-programming/question23.c)
+- [Given two strings X and Y, find the minimum cost to make two strings equal using only delete operations. Cost to delete character in X is S and in Y is P](/dynamic-programming/question24.c)
 
 ## Some important concepts to solve algos better
 

dynamic-programming/question23.c

@@ -0,0 +1,12 @@
+/*
+Longest non-overlapping repeating sub string
+
+Total number of sub strings of a given string are n^2.
+Now, once we have a given string, we need to see whether that substring is a part of the remaining string or
+not. This can be done using KMP algorithm in O(n) time
+Therefore total time complexity: O(n^3)
+
+Better method:
+
+
+*/

dynamic-programming/question24.c

@@ -0,0 +1,103 @@
+/*
+Given two strings X and Y, find the minimum cost to make two strings equal using only delete operations. 
+Cost to delete character in X is S and in Y is P
+
+METHOD1:
+DP using LCS
+Here we need to minimize the number of delete operations, so that the cost can be minimized.
+Hence we need to find as many matching characters in both the strings as possible.
+One way is to find the longest common subsequence in both the strings and the remaining characters
+apart from those should be deleted.
+If the length of one strings is n and other is m
+
+Time complexity: O(mn) //to find LCS
+Space complexity: O(mn)
+
+If length of LCS is l
+Cost of deletion: (m-l)S + (n-l)P
+
+METHOD2:
+Here we use DP:
+C(i,j) = {
+	
+	C(i-1,j-1) if s[i] == s[j],
+	min{ C(i-1,j),(j-1,i) } , if not equal
+
+
+}
+
+Therefore we need to minimize cost. Assume that the length of one string is n and other i m. In worst case
+C(i,j) can have m*n unique subproblems. A binary tree can be made to see the repeating sub problems as per the
+given recurive equation. Hence using bottom-up dynamic programming algo, we can solve this by making a 2d array
+with length 1 greater than the total length of the string.
+Here 0,1 means that we are comparing string with no letter with string with 1st letter and trying to find 
+the answer (0,2): string with no letter with string with two letters and so on
+
+Time complexity: O(mn)
+Space complexity: O(mn)
+*/
+
+//METHOD2
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+int findMin(int a,int b){
+	return (a<b)?a:b;
+}
+
+int findCost(char *str1, char *str2, int cost1, int cost2){
+	int len1 = strlen(str1);
+	int len2 = strlen(str2);
+
+	int sol[len1+1][len2+1];
+
+	int i,j;
+
+	sol[0][0] = 0;
+	
+	for(i=1;i<len1;i++){
+		sol[i][0] = cost1*i;
+	}
+
+	for(j=1;j<len1;j++){
+		sol[0][j] = cost2*j;		
+	}
+
+	for(i=1;i<len1+1;i++){
+		for(j=1;j<len2+1;j++){
+			if(str1[i-1] == str2[j-1]){
+				sol[i][j] = sol[i-1][j-1];
+			}else{
+				sol[i][j] = findMin(sol[i-1][j]+1,sol[i][j-1] + 2);
+			}
+		}
+	}
+
+	printf("priting the sol array\n");
+
+	for(i=0;i<len1+1;i++){
+		for(j=0;j<len2+1;j++){
+			printf("%d  ", sol[i][j]);
+		}
+		printf("\n");
+	}
+
+	return sol[len1][len2];
+}
+
+int main(){
+
+	char str1[] = "RAVI";
+	int cost1 = 1;
+	int cost2 = 2;
+	char str2[] = "AIBD";
+
+	int cost = findCost(str1,str2, cost1, cost2);
+
+	printf("min cost for making two strings equal is: %d\n", cost);
+
+	return 0;
+}
+
+

nextquestions.md

@@ -88,4 +88,7 @@ TODO:
 - program to find fibonacci number in logn time (geeks for geeks)
 - DP question 19 to be done
 - finding all palindromes in a given string
+- generalize the steps in question22 to m steps at a time
+- DP question23 to be done
+