LeetCode in Kotlin

1674. Minimum Moves to Make Array Complementary

Medium

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:

Solution

class Solution {
    fun minMoves(nums: IntArray, limit: Int): Int {
        val delta = IntArray(2 * limit + 2)
        val n = nums.size
        for (i in 0 until n / 2) {
            val a = nums[i]
            val 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]++
        }
        var res = 2 * n
        var curr = 0
        for (i in 2..2 * limit) {
            curr += delta[i]
            res = Math.min(res, curr)
        }
        return res
    }
}