Submit Two-Pointers-2 PR

[agk-s30]

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[super30admin]

Merge Sorted Array (Problem_1.py)

It appears there has been a significant mix-up. You have submitted a solution for "Remove Duplicates from Sorted Array II" instead of "Merge Sorted Array".

To solve "Merge Sorted Array" correctly, you should:

1. Use the two-pointer technique from the end: Start with pointers p1 = m - 1 (last valid element in nums1) and p2 = n - 1 (last element in nums2), and an index idx = m + n - 1 (last position in nums1).
2. Compare and place the larger element at position idx, then decrement the corresponding pointer and idx.
3. Handle remaining elements: After the main loop, if p2 >= 0, copy remaining elements from nums2 to nums1. (No need to handle remaining nums1 elements since they're already in place.)
4. Achieve O(m + n) time and O(1) space by working in-place from the back.

Here's a skeleton to get you started:

class Solution:
    def merge(self, nums1, m, nums2, n):
        p1, p2 = m - 1, n - 1
        idx = m + n - 1
        
        while p1 >= 0 and p2 >= 0:
            if nums2[p2] > nums1[p1]:
                nums1[idx] = nums2[p2]
                p2 -= 1
            else:
                nums1[idx] = nums1[p1]
                p1 -= 1
            idx -= 1
        
        while p2 >= 0:
            nums1[idx] = nums2[p2]
            p2 -= 1
            idx -= 1

Please re-submit with the correct solution for "Merge Sorted Array".

VERDICT: NEEDS_IMPROVEMENT

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Search a 2D Matrix II (Problem_2.py)

VERDICT REASON: [Brief reason for verdict]

VERDICT appears to be NEEDS_IMPROVEMENT

VERDICT: NEEDS_IMPROVEMENT

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Remove Duplicates from Sorted Array II (Problem_3.py)

It appears that you have submitted a solution to the wrong problem. The problem you need to solve is "Remove Duplicates from Sorted Array II," which involves modifying a sorted array in-place so that each unique element appears at most twice. Your submitted code solves "Search a 2D Matrix II" instead.

To correctly solve the assigned problem, you should:

• Use a two-pointer approach (slow and fast pointers) to traverse the sorted array.
• Keep track of how many times the current element has appeared (a count).
• If the count is ≤ 2, copy the element to the slow pointer position and advance the slow pointer.
• Always advance the fast pointer.
• Return the slow pointer as the new length k.

For example, a correct approach would look like:

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        slow = 0
        count = 0
        for fast in range(len(nums)):
            if fast == 0 or nums[fast] != nums[fast - 1]:
                count = 1
            else:
                count += 1
            if count <= 2:
                nums[slow] = nums[fast]
                slow += 1
        return slow

Please re-submit with the correct solution to the assigned problem.

VERDICT: NEEDS_IMPROVEMENT