Completed Backtracking-2

palindrome-partitioning.py

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+''' 
+Time Complexity: O(n * 2^n)
+Space Complexity: O(n)
+
+Approach: We use backtracking to explore all possible partitions of the string. At each step, we check if the current substring is a palindrome. 
+If it is, we include it in the current partition and continue exploring further.
+ If we reach the end of the string, we add the current partition to the result.
+
+'''
+
+class Solution:
+    def partition(self, s: str) -> List[List[str]]:
+
+        path = []
+        res = [] 
+
+        def solve(start):
+            if start>=len(s):
+                res.append(path.copy())
+                return 
+
+            for i in range(start, len(s)):
+                if s[start:i+1]==s[start:i+1][::-1]:
+                    path.append(s[start:i+1])
+                    solve(i+1)
+                    path.pop()
+        solve(0)
+        return res
+

subsets.py

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+'''
+Time Complexity: O(2^n)
+Space Complexity: O(n)
+
+Approach: We use backtracking to explore all possible subsets of the given list. At each step, we have two choices: either include the current element in the subset or exclude it. We explore both choices recursively and add the resulting subsets to the final result.
+
+'''
+class Solution:
+    def subsets(self, nums: List[int]) -> List[List[int]]:
+        N = len(nums)
+        res = []
+        path = []
+        def backtrack(index):
+            if index>=N:
+                res.append(path.copy())
+                return 
+
+            backtrack(index+1)
+            path.append(nums[index])
+            backtrack(index+1)
+            path.pop()
+
+        backtrack(0)
+        return res
+