1674. Minimum Moves to Make Array Complementary

Problem

You are given an integer array nums of even length n and an integer limit. In one move, you can replace any integer from nums with another integer between 1 and limit, inclusive.

The array nums is complementary if for all indices i (0-indexed), nums[i] + nums[n - 1 - i] equals the same number. For example, the array [1,2,3,4] is complementary because for all indices i, nums[i] + nums[n - 1 - i] = 5.

Return the minimum* number of moves required to make nums *complementary.

  Example 1:

Input: nums = [1,2,4,3], limit = 4
Output: 1
Explanation: In 1 move, you can change nums to [1,2,2,3] (underlined elements are changed).
nums[0] + nums[3] = 1 + 3 = 4.
nums[1] + nums[2] = 2 + 2 = 4.
nums[2] + nums[1] = 2 + 2 = 4.
nums[3] + nums[0] = 3 + 1 = 4.
Therefore, nums[i] + nums[n-1-i] = 4 for every i, so nums is complementary.

Example 2:

Input: nums = [1,2,2,1], limit = 2
Output: 2
Explanation: In 2 moves, you can change nums to [2,2,2,2]. You cannot change any number to 3 since 3 > limit.

Example 3:

Input: nums = [1,2,1,2], limit = 2
Output: 0
Explanation: nums is already complementary.

  Constraints:

• n == nums.length

• 2 <= n&nbsp;<=&nbsp;10^5

• 1 <= nums[i]&nbsp;<= limit <=&nbsp;10^5

• n is even.

Solution (Java)

class Solution {
    public int minMoves(int[] nums, int limit) {
        int[] delta = new int[2 * limit + 2];
        int n = nums.length;
        for (int i = 0; i < n / 2; i++) {
            int a = nums[i];
            int b = nums[n - 1 - i];
            delta[2] += 2;
            delta[Math.min(a, b) + 1]--;
            delta[a + b]--;
            delta[a + b + 1]++;
            delta[Math.max(a, b) + limit + 1]++;
        }
        int res = 2 * n;
        int curr = 0;
        for (int i = 2; i <= 2 * limit; i++) {
            curr += delta[i];
            res = Math.min(res, curr);
        }
        return res;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).