chore(CPlusPlus): created anagram (#336)

* Created anagram.cpp

* [Update] : anagram.cpp

* [Fixed] : typo

* [Updated] ; readme.md

algorithms/CPlusPlus/README.md

@@ -57,6 +57,7 @@
 2. [All subsequence of a string (Recursion) ](Strings/sequence.cpp)
 3. [String reversal](Strings/string-reverse.cpp)
 4. [String tokanisation](Strings/string-tokeniser.cpp)
+5. [Anagram check](Strings/anagram.cpp)
 
 ## Trees
 1. [Creating Binary Tree](Trees/build-binary-tree.cpp)

algorithms/CPlusPlus/Strings/anagram.cpp

@@ -0,0 +1,34 @@
+/* 
+   Description : To check whether a string is anagram or not.
+   Anagram: An anagram is a word or phrase formed by rearranging the letters in another word or phrase
+   Approach:Using characters as array index
+*/
+#include<bits/stdc++.h>
+using namespace std;
+
+bool isAnagram(string s1, string s2) {
+    int temp[256] = { 0 };
+    for (int i = 0; i < s1.length(); i++) {         //Idea : Here we are using characters of the string as indexes.
+        temp[s1[i]]++;                              //Each character(except same) will have different position allocated in the temp array      
+        temp[s2[i]]--;                              //Here, characters of first string are incrementing the count at s1[i] position and characters
+                                                    //of second string are decrementing the count at s2[i] position
+    }
+    for (int i = 0; i < 256; i++) {
+        if (temp[i] > 0) {                         //If we found any index in temp array with value greater than 0 then the
+            return false;                          //strings are not anagram
+        }
+    }
+    return true;                                   //If values at all the indexes of temp array is zero, this will indicate that all characters
+}                                                  //in second array cancelled the count of characters in the first array
+int main() {
+    string s1 ;
+    string s2 ;
+    cin>>s1;
+    cin>>s2;
+    string ans = isAnagram(s1, s2) ? "YES" : "NO";
+    cout << ans << endl;
+    return 0;
+    //Ex :  s1= "abc" s2="abd" 
+    //Output : NO
+    //Time Complexity : O(n), where  n is the length of string
+}