diff --git a/groupAnagrams.py b/groupAnagrams.py
new file mode 100644
index 00000000..11cf7988
--- /dev/null
+++ b/groupAnagrams.py
@@ -0,0 +1,54 @@
+# -------- Solution 1 : By sorting the strings------
+''' Time Complexity : O(nklogk) : where n = no of strings in the input
+                                        k = maximum/average length of the string
+                                        O(n) = to process all strings in input, O(klogk) = to sort individual string
+    Space Complexity : O(nk) : 
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+   Your code here along with comments explaining your approach
+
+   Approach : 1. Sort each string in the input 
+             2. create the index of sorted string and Check if it is already mapped in hashmap,
+             3. append the input string to sorted string index
+'''
+
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        hashmap=defaultdict(list)
+        for s in strs:
+            ss = "".join(sorted(s))
+            hashmap[ss].append(s)
+        result=[]
+        for v in hashmap.values():
+            result.append(v)
+        return result
+
+
+# -------- Solution 2 : By hashing using Prime numbers------
+''' Time Complexity : O(nk) : where n = no of strings in the input
+                                        k = maximum/average length of the string
+    Space Complexity : O(nk) : 
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+   Your code here along with comments explaining your approach
+
+   Approach : 1.Create a unique key by multiplying prime nos assigned to each char
+             2. Append the s which has unique keys
+'''
+
+from collections import defaultdict
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        hashmap=defaultdict(list)
+        primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29,
+        31, 37, 41, 43, 47, 53, 59, 61, 67, 71,
+        73, 79, 83, 89, 97, 101 ]
+
+        for s in strs:
+            key = 1
+            for c in s:
+                key *= primes[ord(c) - ord('a')]
+            hashmap[key].append(s)
+        return list(hashmap.values())   
diff --git a/isIsomorphic.py b/isIsomorphic.py
new file mode 100644
index 00000000..7e1f800e
--- /dev/null
+++ b/isIsomorphic.py
@@ -0,0 +1,32 @@
+''' Time Complexity : O(n) , where n = length of the string s or t
+    Space Complexity : O(1) , hashmap will save mapping of max 26 char irrespective of string length, hence constant time.
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+   Your code here along with comments explaining your approach
+
+   Approach : Using two hashmap, one for mapping s to t, and second for mapping t to s
+              Iterating the strings, and checking if the char is previously mapped to some else char if yes, return False
+
+'''
+Class Solution:
+    def isIsomorphic(self, s: str, t: str) -> bool:
+        if len(s) != len(t):
+            return False
+        shash, thash = {}, {}
+        for i in range(len(s)):
+            schar, tchar = s[i],t[i]
+            # s to t mapping
+            if schar in shash:
+                if shash[schar] != tchar:
+                    return False
+            else:
+                shash[schar] = tchar
+                
+            # t to s mapping
+            if tchar in thash:
+                if thash[tchar] != schar:
+                    return False
+            else:
+                thash[tchar] = schar
+        return True    
\ No newline at end of file
diff --git a/word_pattern.py b/word_pattern.py
new file mode 100644
index 00000000..8efa8940
--- /dev/null
+++ b/word_pattern.py
@@ -0,0 +1,40 @@
+''' Time Complexity : O(n) , where n = length of the string s or pattern
+    Space Complexity : O(1) , hashmap will save mapping of max 26 char irrespective of string length, hence constant time.
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+   Your code here along with comments explaining your approach
+
+   Approach : Using two hashmap, one for mapping pattern to s, and second for vice verca
+              Iterating the strings, and checking if the char is previously mapped to some else word if yes, return False
+
+'''
+class Solution:
+    def wordPattern(self, pattern: str, s: str) -> bool:
+        s = s.split()
+        if len(pattern) != len(s):
+            return False
+        # for pattern to s mapping
+
+        phashmap = {}
+        for i in range(len(pattern)):
+            c = pattern[i]
+            if c in phashmap:
+                if phashmap[c] != s[i]:
+                    return False
+            else:
+                phashmap[c] = s[i]
+
+        # for s to pattern mapping
+        shashmap = {}
+        for i in range(len(s)):
+            word = s[i]
+            if word in shashmap:
+                if shashmap[word] != pattern[i]:
+                    return False
+            else:
+                shashmap[word] = pattern[i]
+        return True
+
+
+        
\ No newline at end of file