Backtracking-2

PalindromePartitioning.py

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+# Time Complexity : O(2^n * n) -> For n length string, exponential partitions × (palindrome checks + creating substrings)
+# Space Complexity : O(n^2) (substring creation and recursion stack)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach : We start at the first index and keep trying all substrings from that point.
+# Whenever we find a palindrome substring, we add it and recurse from the next index.
+# Once we reach the end, we save the current list, and then backtrack to try other splits.
+
+class Solution:
+    def partition(self, s: str) -> List[List[str]]:
+        self.res = []
+        self.helper(s, 0, [])
+        return self.res
+
+    def helper(self, s, pivot, path):
+        if pivot == len(s):
+            self.res.append(list(path))
+            return
+
+        for i in range(pivot, len(s)):
+            subStr = s[pivot:i+1]
+            if self.isPalindrome(subStr):
+                path.append(subStr)
+                self.helper(s, i+1, path)
+                path.pop()
+
+    def isPalindrome(self, s):
+        left, right = 0, len(s)-1
+        while left < right:
+            if s[left] != s[right]:
+                return False
+            left += 1
+            right -= 1
+        return True
+

Subsets.py

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+# Time Complexity : O(n × 2^n) - 2^n subsets × up-to-n cost to copy path into the result list
+# Space Complexity : O(n)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach : We iterate from the current pivot and add each element one by one into the path.
+# For every choice, we recurse forward and then backtrack to explore other paths.
+# We save a copy of the current list at every level to record all subsets.
+
+class Solution:
+    def subsets(self, nums: List[int]) -> List[List[int]]:
+        self.res = []
+        self.path = []
+        self.helper(nums, 0)
+        return self.res
+
+    def helper(self, nums, pivot):
+        self.res.append(self.path[:])
+        for i in range(pivot, len(nums)):
+            self.path.append(nums[i])
+            self.helper(nums, i+1)
+            self.path.pop()
+