Fixed typos

Author: ss18

README.md

@@ -76,7 +76,7 @@ Include contains single header implementation of data structures and some algori
 ### Cracking the coding interview problems
 | Problem | Solution |
 | :------------ | :----------: |
-| Problem 1-1 : Edition 6: Write an algorithm to determine whether a string has unique characters or not. Can we do it without using addtional data structures? | [1-1-hasUniqueChars.cpp](cracking_the_coding_interview_problems/1-1-hasUniqueChars.cpp)|
+| Problem 1-1 : Edition 6: Write an algorithm to determine whether a string has unique characters or not. Can we do it without using additional data structures? | [1-1-hasUniqueChars.cpp](cracking_the_coding_interview_problems/1-1-hasUniqueChars.cpp)|
 | Problem 1-2 : Edition 5: Reverse a string when you are a pass a null terminated C string.|[1-2-edi5-reverseString.cpp ](cracking_the_coding_interview_problems/1-2-edi5-reverseString.cpp)|
 | Problem 1-2 : Edition 6: Given two strings, determine if one is permutation of other.|[1-2-perm-strings.cpp](cracking_the_coding_interview_problems/1-2-perm-strings.cpp)|
 | Problem 1-3 : Edition 5: Write an algorithm to remove duplicate chars from a string.|[1-3-edi5-removeDuplicates.cpp](cracking_the_coding_interview_problems/1-3-edi5-removeDuplicates.cpp)|

bit_manipulation/right_most_set_bit.cpp

@@ -1,7 +1,7 @@
 /**
  *  Problem : One line function to return the position of right most bit.
  *  Approach : take 2's compliment and it with number.
- *  And finally taking a log of 2  + 1 will give us the positon
+ *  And finally taking a log of 2  + 1 will give us the position
  */
 
 #include <iostream>

cracking_the_coding_interview_problems/1-1-hasUniqueChars.cpp

@@ -2,7 +2,7 @@
  * Cracking the coding interview, edition 6
  * Problem 1.1
  * Write an algorithm to determine whether a string has unique characters or not
- * Can we do it without using addtional data structures?
+ * Can we do it without using additional data structures?
  */
 
 

cracking_the_coding_interview_problems/1-3-URLify.cpp

@@ -1,7 +1,7 @@
 /*
  * Cracking the coding interview Edition 6
  * Problem 1.3 URLify --> Replace all the spaces in a string with '%20'. 
- * Assumption : We have enough space to accomodate addition chars
+ * Assumption : We have enough space to accommodate addition chars
  * Preferebly in place
  */
 
@@ -10,7 +10,7 @@
 
 /*
  * Function : urlify
- * Args     : string long enough to accomodate extra chars + true len 
+ * Args     : string long enough to accommodate extra chars + true len 
  * Return   : void (in place transformation of string)
  */
 

cracking_the_coding_interview_problems/2-2-kthToLast.cpp

@@ -9,7 +9,7 @@
  *  1. Iterative:
  *  	 Use two pointers
  *  	 Move first pointer k places.
- *  	 Now move second pointer(from start) and first pointer (from k+1) simultanously.
+ *  	 Now move second pointer(from start) and first pointer (from k+1) simultaneously.
  *  	 When second pointer would have reached end, first pointer would be at kth node.
  *
  *  2. Recursive:

leet_code_problems/shortest_path_maze.cpp

@@ -54,7 +54,7 @@ int shortestPath(const std::vector<std::vector<int>>& matrix,
     const Point& source,
     const Point& destination)
 {
-    // An auxillary matrix to keep track of visited points
+    // An auxiliary matrix to keep track of visited points
     // initially all cells are marked unvisited.
     //
     std::vector<std::vector<bool>> visited(

string_problems/robinKarpStringMatching.cpp

@@ -1,5 +1,5 @@
 /*
- * Given a string pattern(P) and large Text string (T), Write a function search( P , T) which provide all the occurances of P in T.
+ * Given a string pattern(P) and large Text string (T), Write a function search( P , T) which provide all the occurrences of P in T.
  * example : T => "AABAACAADAABAAABAA".
  *           P => "AABA"
  * Output : 0, 9, 13 ( all indices of T where pattern string P is starts to match.
@@ -10,7 +10,7 @@
  *         Lets have a hash function --> hash.
  *         Step 1 :We will calculate hash of Pattern P, lets say it is p
  *         Step 2 : Then we will calculate hash of text portion from T[0-->M-1]. lets say  t(0) 
- *         Step 3: if ( p == t(0) ) if they match, add it to list of occurances.
+ *         Step 3: if ( p == t(0) ) if they match, add it to list of occurrences.
  *         Step 4: Go back to step 2, and calculate t(1) i.e hash of T[1-->M] using t(0) in O(1).
  *
  *         The question remains, how do we calculate t(1) from t(0) in O(1), we do it using Horner's rule