diff --git a/anagrams.py b/anagrams.py
new file mode 100644
index 00000000..44299037
--- /dev/null
+++ b/anagrams.py
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+# Time Complexity : O(N * K)
+# Space Complexity : O(N)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach: Use prime-number hashing to uniquely represent each word.
+# Group words by this key and return all hashmap values.
+
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        h_map = {}
+        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]
+        
+        def getCode(s):
+            res = 1
+            for ch in s:
+                res *= primes[ord(ch) - ord('a')]
+            return res
+
+        for word in strs:
+            idx = getCode(word)
+            if idx not in h_map:
+                h_map[idx] = [word]
+            else:
+                h_map[idx].append(word)
+
+        return list(h_map.values())
diff --git a/isomorphic.py b/isomorphic.py
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index 00000000..0f8eee30
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+++ b/isomorphic.py
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+# Time Complexity : O(n)
+# Space Complexity : O(n)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach: Use a hashmap to store character mappings from string `s` to `t`.
+# Iterate through both strings once and validate the one-to-one mapping.
+
+class Solution:
+    def isIsomorphic(self, s: str, t: str) -> bool:
+        key_map = {}
+        value_set = set()
+
+        for i in range(len(s)):
+            if s[i] not in key_map:
+                if t[i] in value_set:
+                    return False
+                key_map[s[i]] = t[i]
+                value_set.add(t[i])
+            else:
+                if key_map[s[i]] != t[i]:
+                    return False
+        return True
diff --git a/wordPattern.py b/wordPattern.py
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index 00000000..b1ebbf02
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+++ b/wordPattern.py
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+# Time Complexity : O(n)
+# Space Complexity : O(n)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach:
+# Split the input string into words. Use a hashmap to map each character in `pattern` to a word.
+# Iterate once through the pattern and validate.
+
+class Solution:
+    def wordPattern(self, pattern: str, s: str) -> bool:
+        words = s.split()
+        if len(words) != len(pattern):
+            return False
+
+        h_map = {}
+        used = set()
+
+        for i in range(len(pattern)):
+            if pattern[i] not in h_map:
+                if words[i] in used:
+                    return False
+                h_map[pattern[i]] = words[i]
+                used.add(words[i])
+            else:
+                if h_map[pattern[i]] != words[i]:
+                    return False
+
+        return True