diff --git a/Problem1.py b/Problem1.py
new file mode 100644
index 00000000..dfc30586
--- /dev/null
+++ b/Problem1.py
@@ -0,0 +1,20 @@
+#Group Anagrams
+# create a map to group the anagrams
+# the key will be the frequency of the alphabets as in anagrams character frequency are same
+# for saving the frequency use the 26 length array, and once you have the freq convert to tupple and use it as key
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        mapOfCharFreq = {}
+        for string in strs:
+            charArr = [0]*26
+            for char in string:
+                index =  ord(char) - 97
+                charArr[index]+=1
+            charTupple = tuple(charArr)
+            if charTupple in mapOfCharFreq:
+                mapOfCharFreq[charTupple].append(string)
+            else:
+                mapOfCharFreq[charTupple] = [string]
+        return list(mapOfCharFreq.values())
+print(groupAnagrams(["eat","tea","tan","ate","nat","bat"]))
+
diff --git a/Problem2.py b/Problem2.py
new file mode 100644
index 00000000..1b67801f
--- /dev/null
+++ b/Problem2.py
@@ -0,0 +1,24 @@
+# Isomorphic Strings
+class Solution:
+    def isIsomorphic(self, s: str, t: str) -> bool:
+        #create map of char in s to t when validating from s to t, to validate same char is not mapped to 2 different char. If yes it's not isomorphic
+        #rplicate same from t to s
+        # for runtiime optimization do it in 1 pass
+
+        sToTMap = {}
+        tToSMap = {}
+        for i in range(0,len(s)):
+            sChar = s[i]
+            tCHar = t[i]
+            if sChar in sToTMap:
+                if sToTMap[sChar] != tCHar:
+                    return False
+            else:
+                sToTMap[sChar] = tCHar
+
+            if tCHar in tToSMap:
+                if tToSMap[tCHar] != sChar:
+                    return False
+            else:
+                tToSMap[tCHar] = sChar
+        return True
diff --git a/Problem3.py b/Problem3.py
new file mode 100644
index 00000000..5f8750d2
--- /dev/null
+++ b/Problem3.py
@@ -0,0 +1,27 @@
+# Word Pattern
+
+# similar to isopmorphic, here characters in patterns are isomorphic to the str. So maintain two way map and validate mapping remains consistent
+class Solution:
+    def wordPattern(self, pattern: str, s: str) -> bool:
+        lengthPattern = len(pattern)
+        arrayS = s.split(" ")
+        lengthS = len(arrayS)
+        if lengthS != lengthPattern:
+            return False
+        charToStrMap = {}
+        strToCharMap = {}
+        for i in range(0,lengthPattern):
+            key = pattern[i]
+            if key in charToStrMap:
+                if charToStrMap[key] != arrayS[i]:
+                    return False
+            else:
+                charToStrMap[key] = arrayS[i]
+
+            key = arrayS[i]
+            if key in strToCharMap:
+                if strToCharMap[key] != pattern[i]:
+                    return False
+            else:
+                strToCharMap[key] = pattern[i]
+        return True
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