new question added

Author: arorarahul

README.md

@@ -179,6 +179,9 @@ nothing but fibonacci series.
 - 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
 - Somtimes if the problem is already a table, you need to reverse engineer it to find the min number of ways to do something. Example create a new table and start from the very bottom and construct the solution updwards. Answer in that case will be the cell 0,0
+- If a string and its reversal have a common subsequence, then definately that common subsequence is going to be a palindrome
+- Whenever a problem is given involving string, check if LCS can be used
+- Whenever a generic problem is given check if it is becoming a fibonacci series or a form of it.
 
 # Topic0: Programming Questions
 
@@ -403,6 +406,7 @@ it gets converted to i-1,j-1 or we take min or max of i,j-1 and i-1,j
 - [Given an amount and some coints, find out in how many ways can this amount be formed](/dynamic-programming/question26.c)
 - [Given a 2xn board and tiles of size 2x1, count the number of ways to fill the board using 2x1 tiles](/dynamic-programming/question27.c)
 - [Given a cost matrix mxn having a cost at each cell. Find the min cost that it will take to reach cell (m-1,n-1) from top left corner cell (0,0) if the only allowed directions to move are right, down and diagnal down](/dynamic-programming/question28.c)
+- [Given a string, find out if it becomes a palindrome or not after doing atmost k deletions](/dynamic-programming/question29.c)
 
 ## Some important concepts to solve algos better
 

dynamic-programming/question29.c

@@ -0,0 +1,90 @@
+/*
+Given a string, find out if it becomes a palindrome or not after doing atmost k deletions
+
+Naive approach:
+lets say the string is abxca
+and value of k=2;
+then answer is YES, because we can do a min of 1 deletions and max of 2 deletions
+
+Therefore, number of ways to solve this will be
+nc1+ nc2....nck = n^k
+which is exponential
+
+METHOD1:
+DP Longest common subsequence:
+If we reverse the given string and try to find the longest common subsequence of the given string with its
+reversal, definately the LCS is going to be palindrome.
+If the length of the main string and the LCS is subtracted we will get the min number of deletions
+required.
+If that min number is less than the max value of k given in the question answer will be YES else NO.
+
+Time complexity: O(n^2) //for LCS
+Space complexity: O(n^2) //for LCS
+
+Also, space complexity can be reduced to O(n) time if just one row is maintained at any point of time for LCS
+*/
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+void revString(char *str, char *reverse, int len){
+	int i,j;
+	int temp;
+	i = len-1;
+	j = 0;
+	while(i>-0){
+		reverse[j] = str[i];
+		j++;
+		i--;
+	}
+	printf("inside reverse function\n");
+	printf("%s\n", reverse);
+}
+
+int max(int a, int b){
+	return a>b?a:b;
+}
+
+int checkPal(char *str, int k){
+	int len = strlen(str);
+	char reverse[len];
+	revString(str, reverse, len);
+
+	int lcs[len+1][len+1];
+	int i,j;
+	for(i=0;i<=len;i++){
+		lcs[i][0] = 0;
+		lcs[0][i] = 0;
+	}
+
+	for(i=1;i<=len;i++){
+		for(j=1;j<=len;j++){
+			if(str[i-1] == reverse[i-1]){
+				lcs[i][j] = lcs[i-1][j-1] + 1;	
+			}else{
+				lcs[i][j] = max(lcs[i-1][j],lcs[i][j-1]);
+			}
+		}
+	}
+
+	if(lcs[len][len] > k){
+		return 0;
+	}
+	return 1;
+}
+
+int main(){
+	int k;
+	printf("enter the value of k\n");
+	scanf("%d",&k);
+	char str[] = "acbxcda";
+	
+	int isPalindrome = checkPal(str,k);
+	if(isPalindrome){
+		printf("YES\n");
+	}else{
+		printf("NO\n");
+	}
+
+	return 0;
+}

nextquestions.md

@@ -98,5 +98,6 @@ TODO:
 - DP question26 second case
 - Fibonacci series indepth analysis
 - DP question27 variation where size of board is nxm and tile is mx1
+- longest bitonic sequence