Minimum Moves to Make Array Complementary LeetCode solution in Python

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class Solution:
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    def minMoves(self, nums: List[int], limit: int) -> int:
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        n = len(nums)
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        overlay_arr = [0] * (2 * limit + 2)
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        for i in range(n // 2):
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            left_boundary = min(nums[i], nums[n - 1 - i]) + 1
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            no_move_value = nums[i] + nums[n - 1 - i]
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            right_boundary = max(nums[i], nums[n - 1 - i]) + limit
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            overlay_arr[left_boundary] -= 1
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            overlay_arr[no_move_value] -= 1
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            overlay_arr[no_move_value + 1] += 1
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            overlay_arr[right_boundary + 1] += 1
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        curr_moves = n  # initial assumption of two moves for each pair
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        res = float("inf")
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        # start Sweeping
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        for i in range(2, 2 * limit + 1):
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            curr_moves += overlay_arr[i]
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            res = min(res, curr_moves)
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        return res