diff --git a/palindrome-partitioning.py b/palindrome-partitioning.py
new file mode 100644
index 00000000..5d1cff5f
--- /dev/null
+++ b/palindrome-partitioning.py
@@ -0,0 +1,38 @@
+#--------Solution 1 : For loop based recursion------------
+''' Time Complexity : O(n* 2^n)) ;  n = len of the list and O(n) to palindrome check
+    Space Complexity : O(n^2)  : recursion stack + substring creation
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+'''
+
+class Solution:
+    def partition(self, s: str) -> List[List[str]]:
+        self.result = []
+
+        def helper(s, pivot, path):
+            #base
+            if pivot == len(s):
+                self.result.append(list(path))
+
+            #logic
+            for i in range(pivot,len(s)):
+                sub = s[pivot:i+1]
+                if self.isPalindrome(sub):
+                    #action
+                    path.append(sub)
+                    #recurse
+                    helper(s,i+1,path)
+                    #backtrack
+                    path.pop()
+
+        helper(s,0,[])
+        return self.result
+    
+    def isPalindrome(self, sub):
+        l, r = 0, len(sub)-1
+        while l < r:
+            if sub[l] != sub[r]:
+                return False
+            l += 1
+            r -= 1
+        return True
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diff --git a/subsets.py b/subsets.py
new file mode 100644
index 00000000..3c9c84a8
--- /dev/null
+++ b/subsets.py
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+#--------Solution 1 : Backtraking------------
+''' Time Complexity : O(n* 2^n)) ;  n = len of the list and O(n) to copy path to result
+    Space Complexity : O(n) 
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+'''
+class Solution:
+    def subsets(self, nums: List[int]) -> List[List[int]]:
+        self.result = []
+
+        def helper(nums, idx, path):
+            #base
+            if idx == len(nums):
+                self.result.append(list(path))
+                return
+            #logic
+            #no choose
+            helper(nums, idx+1, path)
+
+            #choose
+            path.append(nums[idx])
+            helper(nums, idx+1, path)
+            #backtrack
+            path.pop()
+        
+        helper(nums,0,[])
+        return self.result
+
+#--------Solution 2 : Nested For loop------------
+''' Time Complexity : O(n * n^2)) ;  
+    Space Complexity : O(n) 
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+'''
+class Solution:
+    def subsets(self, nums: List[int]) -> List[List[int]]:
+        result =[[]]
+
+        for i in range(len(nums)):
+            for j in range(len(result)):
+                li = result[j].copy()
+                li.append(nums[i])
+                result.append(li)
+        return result
+
+        
+        
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