Solution for Anagrams problem (different approach)

coding_freshmen/C++/Mahi_Chrungoo_MAHIC1201/ANAGRAMS.cpp

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+#include<bits/stdc++.h>
+using namespace std;
+bool isAnagram(string s1, string s2)
+{
+    sort(s1.begin(),s1.end());
+    sort(s2.begin(),s2.end());
+    if(s1==s2)
+    {
+        return true;
+    }
+    return false;
+}
+int main()
+{
+  bool ans;
+  string a;
+  string b;
+  cin>>a>>b;
+  ans=isAnagram(a,b);
+  if(a.length()!=b.length())
+  {
+    cout<<"false";
+  }
+  else if(ans==1)
+  {
+    cout<<"true";
+  }
+  else if(ans==0)
+  {
+    cout<<"false";
+  }
+}

coding_freshmen/C++/Mahi_Chrungoo_MAHIC1201/Readme.md

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+# Anagram Checker
+
+## Problem Explanation 🚀
+You are given two strings, `s` and `t`. Your task is to determine if `t` is an anagram of `s`. An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.
+
+## Your Logic 🤯
+### Approach
+This problem can be solved by 
+1. Sorting both the strings (lets say s1 and s2). Wr firstly check if the two strings have same/different length, only strings if same length can be anagrams.
+2. Checking if both the strings are equal(using isAnagram() function that takes input of two strings).
+Sorting in ascending order(sorting will be based on ascii values of the characters in the strings).
+3. If both the strings are anagrams,both will get sorted in the same order.
+(eg "anagram" will get sorted to "aaagmnr" and "nagaram" will also get sorted to "aaagmnr", hence they are anagrams), hence return true. 
+4. Strings such as "rat", on sorting will become "atr", whereas "car" on sorting will become "acr", since atr!=acr, both the strings are not anagrams, hence return false.
+
+### Code Structure and Libraries Used
+```cpp
+#include<bits/stdc++.h>
+using namespace std;
+bool isAnagram(string s1, string s2)
+{
+    sort(s1.begin(),s1.end());
+    sort(s2.begin(),s2.end());
+    if(s1==s2)
+    {
+        return true;
+    }
+    return false;
+}
+int main()
+{
+  bool ans;
+  string a;
+  string b;
+  cin>>a>>b;
+  ans=isAnagram(a,b);
+  if(a.length()!=b.length())
+  {
+    cout<<"false";
+  }
+  else if(ans==1)
+  {
+    cout<<"true";
+  }
+  else if(ans==0)
+  {
+    cout<<"false";
+  }
+}
+```
+
+## Time Complexity and Space Complexity
+```cpp
+Example 
+
+Time Complexity -> O(n*log(n)) for sorting
+Space Complexity -> O(1) 
+```
+The time complexity of the solution is O(n * log(n)) because that is the time taken for the sort operation. The space complexity is O(1) because the space used by the two strings a and b created for taking input and a bool variable ans for output.