Skip to content

Backtracking-2 Done #1248

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Open
vaishnavi2231 wants to merge 2 commits into
base: master
Choose a base branch
from
Open

Backtracking-2 Done #1248

Show file tree
Hide file tree
Changes from all commits
Commits
Show all changes
2 commits
Select commit Hold shift + click to select a range
e584bed
Backtracking-2 Done
Feb 1, 2026
7075aed
Delete .DS_Store
vaishnavi2231 Feb 1, 2026
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
38 changes: 38 additions & 0 deletions palindrome-partitioning.py
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,38 @@
#--------Solution 1 : For loop based recursion------------
''' Time Complexity : O(n* 2^n)) ; n = len of the list and O(n) to palindrome check
Space Complexity : O(n^2) : recursion stack + substring creation
Did this code successfully run on Leetcode : Yes
Any problem you faced while coding this : No
'''

class Solution:
def partition(self, s: str) -> List[List[str]]:
self.result = []

def helper(s, pivot, path):
#base
if pivot == len(s):
self.result.append(list(path))

#logic
for i in range(pivot,len(s)):
sub = s[pivot:i+1]
if self.isPalindrome(sub):
#action
path.append(sub)
#recurse
helper(s,i+1,path)
#backtrack
path.pop()

helper(s,0,[])
return self.result

def isPalindrome(self, sub):
l, r = 0, len(sub)-1
while l < r:
if sub[l] != sub[r]:
return False
l += 1
r -= 1
return True
47 changes: 47 additions & 0 deletions subsets.py
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,47 @@
#--------Solution 1 : Backtraking------------
''' Time Complexity : O(n* 2^n)) ; n = len of the list and O(n) to copy path to result
Space Complexity : O(n)
Did this code successfully run on Leetcode : Yes
Any problem you faced while coding this : No
'''
class Solution:
def subsets(self, nums: List[int]) -> List[List[int]]:
self.result = []

def helper(nums, idx, path):
#base
if idx == len(nums):
self.result.append(list(path))
return
#logic
#no choose
helper(nums, idx+1, path)

#choose
path.append(nums[idx])
helper(nums, idx+1, path)
#backtrack
path.pop()

helper(nums,0,[])
return self.result

#--------Solution 2 : Nested For loop------------
''' Time Complexity : O(n * n^2)) ;
Space Complexity : O(n)
Did this code successfully run on Leetcode : Yes
Any problem you faced while coding this : No
'''
class Solution:
def subsets(self, nums: List[int]) -> List[List[int]]:
result =[[]]

for i in range(len(nums)):
for j in range(len(result)):
li = result[j].copy()
li.append(nums[i])
result.append(li)
return result