Backtracking 2 Done

palindrome.py

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+# Time Complexity : O(N.2^N) where N is the size of the list of candidates
+# Space Complexity : O(N.2^N) where N is the size of the recursion stack in the worst case
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+# Your code here along with comments explaining your approach:
+# I am creating an ans list to store the palindromic partitions.
+# I am using a helper function to perform backtracking.
+# The helper function takes the current path and the starting index as input.
+# If the starting index is equal to the length of the string, I append the current path to the ans list.
+# I iterate through the string starting from the starting index to avoid duplicates.
+# For each substring, I check if it is a palindrome using the palindrome helper function.
+# If it is, I append it to the current path and call the helper function recursively with the updated path and the next index.
+# After the recursive call, I remove the last substring from the current path to backtrack.
+# Finally, I return the ans list.
+
+from typing import List
+class Solution:
+    def partition(self, s: str) -> List[List[str]]:
+        ans = []
+        def backtrack(path,idx):
+            if len(s) == idx:
+                ans.append(path[:])
+                return 
+            for i in range(idx,len(s)):
+                if palindrome(s,idx,i):   
+                    path.append(s[idx:i+1]) 
+                    backtrack(path,i+1)
+                    path.pop()
+        def palindrome(path,start,end):
+            while start <= end:
+                if s[start] != s[end]:
+                    return False
+                start += 1
+                end -= 1
+            return True
+        backtrack([],0)
+        return ans
+
+
+
+

subset.py

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+# Time Complexity : O(N.2^N) where N is the size of the list of nums
+# Space Complexity : O(N.2^N) where N is the size of the recursion stack in the worst case
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+# Your code here along with comments explaining your approach:
+# I am creating an ans list to store the subsets.
+# I am using a helper function to perform backtracking.
+# The helper function takes the current path and the starting index as input.
+# I append the current path to the ans list.
+# I iterate through the nums list starting from the starting index to avoid duplicates.
+# For each number, I append it to the current path and call the helper function recursively with the updated path and the next index.
+# After the recursive call, I remove the last number from the current path to backtrack.
+# Finally, I return the ans list.
+
+from typing import List
+class Solution:
+    def subsets(self, nums: List[int]) -> List[List[int]]:
+        ans = []
+        def helper(path,idx):
+            ans.append(path[:])
+            for i in range(idx,len(nums)):
+                path.append(nums[i])
+                helper(path,i+1)
+                path.pop()
+        helper([],0)
+        return ans
+
+