new question added

Author: arorarahul

README.md

@@ -105,6 +105,8 @@ start and end of array is chosen(specifically where array is split again and aga
 - In case of string arr size should be measured using strlen and not using sizeof as sizeof also includes \0
 - Sometimes a string can be merged with itself or the other to solve some algos. For eg: if one string is rotation of the other, concatenating one with itself can give a string where second string will be a substring in this string.
 - Store as many things as required to solve the algo in hash table as it is a structure 
+- It is always good to free memory assigned to hashTable after the end of the program
+- Sometimes hashtable value can be decremented and not incremented to solve an algo for eg. finding anagram
 
 # Topic0: Programming Questions
 
@@ -264,7 +266,8 @@ start and end of array is chosen(specifically where array is split again and aga
 - [Reverse a given string](/strings/question5.c)
 - [Check whether given string is palindrome or not](/strings/question6.c)
 - [Find the first non-repeating character in a given string](/strings/question7.c)
-- [](/strings/question8.c)
+- [Run length encoding](/strings/question8.c)
+- [Check whether given two strings are anagrams of each other](/strings/question9.c)
 
 
 ## Some important concepts to solve algos better
@@ -337,6 +340,11 @@ trees having random number of children not necessary equal
 are at same level but do not have same parent
 - O(1) means time complexity or space complexity is not dependent on the input size
 - One string is rotation of the other, if one of the rotations of one string matches the other one.
+- Run length encoding means, traversing through the given character array and displaying which character
+is repeating how many times along with the character as output. 
+Eg: SSMMMAAARRT => S2M3A3R2T1
+- Two strings are anagrams if they have same no of characters and they are composed of the same letters.
+Even if there are repetitions, they are going to be same. 
 
 # C Programming - Topic1: Introduction
 

nextquestions.md

@@ -57,3 +57,4 @@ TODO:
 - question19 method2 to be done later after strings (revisit the code for it given as pdf)
 - method1 of question21 and question22 complete
 - do binary search method for strings question1
+- question 8 strings to be done (see from the sol pdf)

strings/a.exe

@@ -0,0 +1,41 @@
+/*
+Run length encoding
+
+Run length encoding means, traversing through the given character array and displaying which character
+is repeating how many times along with the character as output. 
+Eg: SSMMMAAARRT => S2M3A3R2T1
+
+Applications: File compression
+
+METHOD:
+Just traverse and maintain a counter to solve the algo
+Time complexity: O(n)
+Space complexity: O(1)
+*/
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+void runLengthEncoding(char *arr, int size){
+	int i,j=0;
+	// char *finalStr = (char *)malloc(sizeof(char)*(size*2 + 1));
+	int counter = 1;
+	printf("%s\n", arr);
+	for(i=0;i<size;i++){
+		// finalStr[j++] = arr[i];
+		if(arr[i] == arr[i+1]){
+			counter++;
+		}else{
+			printf("%c%d", arr[i],counter);
+			counter=1;
+		}
+	}
+}
+
+int main(){
+	char arr[] = "SSMMMAAARRT";
+	int length = strlen(arr);
+
+	runLengthEncoding(arr,length);
+	return 0;
+}

strings/question8.c

@@ -0,0 +1,54 @@
+/*
+Check whether given two strings are anagrams of each other
+
+Two strings are anagrams if they have same no of characters and they are composed of the same letters.
+Even if there are repetitions, they are going to be same. 
+
+METHOD1:
+sort and scan
+Time complexity: O(nlogn)
+Space complexity: O(1) //for inplace sorting
+
+METHOD2: 
+use hash table and increment count for first string and decrement for second. In the end all count
+values should be zero.
+Time complexity: O(n)
+Space complexity: O(1) //256 not dependent on input
+*/
+//sorting method not done as it is very basic
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+#define size 256
+
+void checkAnagram(char *str1, char *str2, int l1, int l2){
+	int hash[size] = {0};
+	int i;
+	for(i=0;i<l1;i++){
+		hash[str1[i]]++;
+	}	
+	for(i=0;i<l2;i++){
+		hash[str2[i]]--;
+	}
+	for(i=0;i<size;i++){
+		if(hash[i] > 0){
+			printf("two strings are NOT anagrams\n");
+			return;
+		}
+	}
+	printf("two strings are anagrams of each other\n");
+}
+
+int main(){
+	char str1[] = "heater";
+	char str2[] = "reheaa";
+	int l1 = strlen(str1);
+	int l2 = strlen(str2);
+	if(l1 != l2){
+		printf("two strings not NOT anagrams\n");
+		return 0;
+	}
+	checkAnagram(str1,str2, l1, l2);
+	return 0;
+}