From 625dc1ce9697e024aeec049eedb09fd0c0d0bd03 Mon Sep 17 00:00:00 2001
From: Shubham Jain <shubham@Mac.lan>
Date: Fri, 30 Jan 2026 17:57:26 -0600
Subject: [PATCH] Completed

---
 Solutions.py | 65 ++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 65 insertions(+)
 create mode 100644 Solutions.py

diff --git a/Solutions.py b/Solutions.py
new file mode 100644
index 00000000..7ad42701
--- /dev/null
+++ b/Solutions.py
@@ -0,0 +1,65 @@
+#Problem1 Subsets (https://leetcode.com/problems/subsets/)
+# Time Complexity : O(N 2^N)
+# Space Complexity : O(N 2^N)
+
+
+
+class Solution(object):
+    def subsets(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: List[List[int]]
+        """
+        
+        def __helper(pivot,res,path):
+            res.append(path[:]) ## append deep copy
+
+            for i in range(pivot,len(nums)):
+                ## choose
+                path.append(nums[i])
+                ## recurse
+                __helper(i+1,res,path)
+                ## no choose
+                path.pop()
+
+        res = []
+        path = []
+        __helper(0,res,path)
+        return res
+
+
+#Problem2 Palindrome Partitioning(https://leetcode.com/problems/palindrome-partitioning/)
+# Time Complexity : O(N 2^N)
+# Space Complexity : O(N 2^N)
+
+class Solution(object):
+    def partition(self, s):
+
+    
+        # Backtracking approach:
+        # - Try every possible substring starting from current index (pivot)
+        # - If the substring is a palindrome, choose it, recurse for the rest
+        # - Backtrack and explore other options
+        def isPalindrome(sub):
+            """Check if a substring is a palindrome."""
+            return sub == sub[::-1]
+
+        def __helper(s, pivot, path, result):
+        
+            # Base case: reached end of string
+            if pivot == len(s):
+                result.append(path[:])  # make a copy of path
+                return
+            
+            # Try all possible substrings starting at pivot
+            for end in range(pivot, len(s)):
+                sub = s[pivot:end+1]  # substring s[pivot:end]
+                if isPalindrome(sub):
+                    path.append(sub)            # choose
+                    __helper(s, end+1, path, result)  # recurse
+                    path.pop()                  # backtrack
+
+        result = []
+        path = []
+        __helper(s, 0, path, result)
+        return result