Done Backtracking-2

[pranjay01]

No description provided.

[super30admin]

Subsets (Problem1.py)

Your solution is correct and efficient, but there are a few areas for improvement:

1. Avoid using class variables for result and tmpResult to prevent state issues across multiple calls. Instead, define them within the subsets method and pass them as arguments or use local variables in the recursive function.
2. Consider using more descriptive variable names and comments. For example, instead of tmpResult, you could use current_subset. Also, the comments "no choose" and "choose" could be clarified to "exclude current element" and "include current element".
3. Remove the test code at the bottom if this is intended for a platform like LeetCode, as it may cause runtime errors.
4. You can make the helper function a nested function to avoid passing nums repeatedly and to encapsulate the recursion.

Here's a revised version of your code that addresses these points:

from typing import List

class Solution:
    def subsets(self, nums: List[int]) -> List[List[int]]:
        result = []
        current_subset = []
        
        def backtrack(index):
            if index == len(nums):
                result.append(current_subset[:])
                return
            
            # Exclude the current element
            backtrack(index + 1)
            
            # Include the current element
            current_subset.append(nums[index])
            backtrack(index + 1)
            current_subset.pop()
        
        backtrack(0)
        return result

This version uses local variables and a nested function, which is cleaner and avoids potential state issues.

VERDICT: PASS

---

Palindrome Partitioning (Problem2.py)

Strengths:

• The solution correctly implements the backtracking algorithm.
• The code is structured with helper functions, which improves readability.
• The use of backtracking (appending and popping) is correctly implemented.

Areas for Improvement:

1. Palindrome Check: Instead of using a recursive function for palindrome check, use an iterative method. This avoids recursion overhead and potential stack issues (even though the string length is limited to 16, it's good practice). For example:
   def isPalindrome(s, start, end):
   while start < end:
   if s[start] != s[end]:
   return False
   start += 1
   end -= 1
   return True

   Then call it with: if isPalindrome(s, pivot, i):

2. Substring Creation: Currently, you create a substring each time with s[pivot:i+1]. This creates a new string and has O(n) time and space. Instead, you can avoid creating the substring until necessary (only when adding to the partition). However, since you need to add the substring to the partition list, you have to create it. Alternatively, you could store indices and only create the substring when adding to the result, but that might complicate the code. Given the constraints, it's acceptable.

3. Variable Names: Consider using more descriptive variable names. For example, pivot is clear, but partitionsList could be current_partition or path (as in the reference). Also, startIndex and endIndex in the palindrome function are clear.

4. Edge Cases: Your solution handles the base case correctly when pivot reaches the end. Also, the recursive palindrome check correctly returns True for empty or single-character strings (via the base condition). However, the iterative palindrome check would be more efficient.

5. Type Hints: You have used type hints for the main function, which is good. However, the helper functions could also benefit from type hints for clarity.

Revised Code Example:
Here's a revised version with iterative palindrome check and minor improvements:

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

    def is_palindrome(start, end):
        while start < end:
            if s[start] != s[end]:
                return False
            start += 1
            end -= 1
        return True

    def helper(pivot, path):
        if pivot == len(s):
            result.append(path[:])
            return
        for i in range(pivot, len(s)):
            if is_palindrome(pivot, i):
                path.append(s[pivot:i+1])
                helper(i+1, path)
                path.pop()

    helper(0, [])
    return result

This version uses an iterative palindrome check and renames variables for clarity.

VERDICT: PASS