Add solutions for palindrome partitioning and subsets

PalindromePartioning.py

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+# Time Complexity : O(n. 2^n)
+# Space Complexity : O(n^2)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+# The approach here to do a backtrack using the iterative for loop approach. But before continuing the recursion, check if the
+# substring is a palindrome or not.
+
+class Solution:
+    def partition(self, s: str) -> List[List[str]]:
+        result = []
+
+        def backtrack(pivot, path):
+            # Base case
+            if pivot == len(s):
+                result.append(path.copy())
+                return
+            # Logic
+            for i in range(pivot, len(s)):
+                subString = s[pivot: i + 1]
+                if isPalindrome(subString):
+                    path.append(subString)
+                    backtrack(i + 1, path)
+                    path.pop()
+
+        def isPalindrome(currStr):
+            p1 = 0
+            p2 = len(currStr) - 1
+
+            while (p1 <= p2):
+                if currStr[p1] != currStr[p2]:
+                    return False
+                p1 += 1
+                p2 -= 1
+            return True
+
+        backtrack(0, [])
+        return result
+

subsets.py

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+# Time Complexity : O(2^n)
+# Space Complexity : O(n^2)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+# The approach here to do a backtrack either using 01 approach(choose, not choose) or using the iterative for loop approach to
+# get all the subsets.
+
+class Solution:
+    def subsets01Recursion(self, nums: List[int]) -> List[List[int]]:
+        result = []
+
+        def helper(path, idx):
+            if idx == len(nums):
+                result.append(path.copy())
+                return
+
+            helper(path, idx + 1)
+            path.append(nums[idx])
+            helper(path, idx + 1)
+            path.pop()
+
+        helper([], 0)
+
+        return result
+
+
+class Solution:
+    def subsetsIterativeRecursion(self, nums: List[int]) -> List[List[int]]:
+        result = []
+
+        def helper(path, pivot):
+            result.append(path.copy())
+            for i in range(pivot, len(nums)):
+                path.append(nums[i])
+                helper(path, i + 1)
+                path.pop()
+
+        helper([], 0)
+
+        return result
+