Done Two pointers 1

[pranjay01]

No description provided.

[super30admin]

Strengths:

• You have implemented the Dutch National Flag algorithm correctly, which is the standard approach for this problem.
• The time and space complexity are optimal.
• The code is well-structured and the variable names are meaningful.

Areas for Improvement:

• Avoid retrieving values from the array into temporary variables when not necessary. For example, in the swap operations, you can directly use the array indices without storing them in variables. This reduces the number of operations and makes the code cleaner.
• The comment "move right pointer by 1 on left" is unclear. It should be clarified that when nums[midPointer] is 2, we swap with the rightPointer and then decrement the rightPointer.
• In the case when nums[midPointer] is 0, after swapping with leftPointer, you increment both leftPointer and midPointer. This is correct, but note that after swapping, the value at midPointer is now 1 (since the leftPointer was pointing to a value that has already been processed), so it's safe to increment midPointer. However, it would be good to explain this in comments for clarity.
• Consider adding a helper function for swapping to make the code more modular and readable, similar to the reference solution.

Suggested improvements for Problem1.py:

class Solution:
    def sortColors(self, nums: List[int]) -> None:
        left = 0
        mid = 0
        right = len(nums) - 1
        
        while mid <= right:
            if nums[mid] == 2:
                nums[mid], nums[right] = nums[right], nums[mid]
                right -= 1
            elif nums[mid] == 0:
                nums[mid], nums[left] = nums[left], nums[mid]
                left += 1
                mid += 1
            else:
                mid += 1

This version is more concise and avoids unnecessary variables.