Working Solution

[avcode3]

No description provided.

[super30admin]

Your solution for the "Merging of 2 arrays" problem is correct and efficient. Well done! Here are a few suggestions to improve code clarity and adhere to the problem requirements:

• Instead of modifying the parameters m and n (which are originally the counts of elements), it is better to create new variables for the pointers. For example, you can use p1 = m - 1 and p2 = n - 1 to point to the last valid elements in nums1 and nums2, respectively. This makes the code more readable and avoids changing the meaning of the parameters.

• The problem states that the function should not return anything. Your solution returns nums1, which is not necessary since the array is modified in-place. You should remove the return statement to match the problem's requirement.

• Consider adding a comment to explain the algorithm briefly, which can help others understand your approach.

Here is a slightly revised version of your code with these changes:

class Solution:
    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
        p1 = m - 1  # pointer for the last element in the initial part of nums1
        p2 = n - 1  # pointer for the last element in nums2
        idx = m + n - 1  # pointer for the last position in nums1

        while p1 >= 0 and p2 >= 0:
            if nums1[p1] >= nums2[p2]:
                nums1[idx] = nums1[p1]
                p1 -= 1
            else:
                nums1[idx] = nums2[p2]
                p2 -= 1
            idx -= 1

        # If there are remaining elements in nums2, copy them
        while p2 >= 0:
            nums1[idx] = nums2[p2]
            p2 -= 1
            idx -= 1

Overall, your solution is excellent. Keep up the good work!