Done Backtracking-2

Problem1.py

@@ -0,0 +1,33 @@
+#Subsets
+
+# use recrussion to create the possible subsets, use tmp result and keep on building on that for every recrussion and once crossed pop it
+# Time -> N2^N rrecrusion branching 2 options at step and then building the final result 
+# Space -> ON the tmpResult and the stack space
+
+from typing import List
+
+
+class Solution:
+    def subsets(self, nums: List[int]) -> List[List[int]]:
+        self.result = []
+
+        self.tmpResult = []
+
+        self.helper(0, nums)
+        return self.result
+
+    def helper(self,index, nums):
+
+        if index == len(nums):
+            self.result.append(self.tmpResult[:])
+            return
+        #no choose 
+        self.helper(index+1, nums)
+
+        #choose
+        self.tmpResult.append(nums[index])
+        self.helper(index+1, nums)
+        self.tmpResult.pop()
+
+sol = Solution()
+print(sol.subsets([1,2,3]))

Problem2.py

@@ -0,0 +1,25 @@
+# Apace  -> On
+# Time NO(2^N)
+# Logic -> use helper and pivot to find all the partitions and validate if those are palindrome or not
+
+class Solution:
+    def partition(self, s: str) -> List[List[str]]:
+        result = []
+
+        def isPalindrome(startIndex, endIndex):
+            if startIndex >= endIndex: return True
+            return s[startIndex] == s[endIndex] and isPalindrome(startIndex+1, endIndex-1)
+
+        def helper(pivot, partitionsList):
+            if pivot == len(s):
+                result.append(partitionsList[:])
+                return
+            for i in range(pivot, len(s)):
+                if not isPalindrome(pivot, i):
+                    continue
+                subString = s[pivot:i+1]                
+                partitionsList.append(subString)
+                helper(i+1, partitionsList)
+                partitionsList.pop()
+        helper(0, [])
+        return result