From 2b7bfb12a2d7d2bc8e85f48ca5e356b8ccb5e89c Mon Sep 17 00:00:00 2001
From: Srijha-Kalyan <87617310+Srijha-Kalyan@users.noreply.github.com>
Date: Mon, 8 Dec 2025 20:47:42 -0500
Subject: [PATCH] Add files via upload

---
 Problem1_subsets.py                 | 22 ++++++++++++++++++++++
 Problem2_palindrome_partitioning.py | 29 +++++++++++++++++++++++++++++
 2 files changed, 51 insertions(+)
 create mode 100644 Problem1_subsets.py
 create mode 100644 Problem2_palindrome_partitioning.py

diff --git a/Problem1_subsets.py b/Problem1_subsets.py
new file mode 100644
index 00000000..9344e61a
--- /dev/null
+++ b/Problem1_subsets.py
@@ -0,0 +1,22 @@
+class Solution(object):
+    def subsets(self, nums):
+        """
+        Time Complexity: O(N * 2^N) where N is the length of nums
+        Space Complexity: O(N) for the recursion stack
+        :type nums: List[int]
+        :rtype: List[List[int]]
+        """
+        res = []
+
+        def backtrack(start, path):
+            #No base case
+            res.append(path[:])
+
+            #Logic
+            for i in range(start, len(nums)):
+                path.append(nums[i])
+                backtrack(i+1, path)
+                path.pop()
+        
+        backtrack(0, [])
+        return res
\ No newline at end of file
diff --git a/Problem2_palindrome_partitioning.py b/Problem2_palindrome_partitioning.py
new file mode 100644
index 00000000..cc2a7d8b
--- /dev/null
+++ b/Problem2_palindrome_partitioning.py
@@ -0,0 +1,29 @@
+class Solution(object):
+    def partition(self, s):
+        """
+        Time Complexity: O(N * 2^N) where N is the length of the string s.
+        Space Complexity: O(N) for the recursion stack and path storage.
+        :type s: str
+        :rtype: List[List[str]]
+        """
+        res = []
+
+        def isPalindrome(substr):
+            return substr==substr[::-1]
+
+        def backtrack(start, path):
+            #base case
+            if start==len(s):
+                res.append(path[:])
+                return
+
+            #logic
+            for i in range(start+1, len(s)+1):
+                substr = s[start:i]
+                if isPalindrome(substr):
+                    path.append(substr)
+                    backtrack(i, path)
+                    path.pop()
+        
+        backtrack(0, [])
+        return res
\ No newline at end of file