花花酱 LeetCode 1674. Minimum Moves to Make Array Complementary

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 <= 105
• 1 <= nums[i] <= limit <= 105
• n is even.

Solution: Sweep Line / Prefix Sum

Let a = min(nums[i], nums[n-i-1]), b = max(nums[i], nums[n-i-1])
The key to this problem is how many moves do we need to make a + b == T.

if 2 <= T < a + 1, two moves, lower both a and b.
if a +1 <= T < a + b, one move, lower b
if a + b == T, zero move
if a + b + 1 <= T < b + limit + 1, one move, increase a
if b + limit + 1 <= T <= 2*limit, two moves, increase both a and b.

Time complexity: O(n + limit) or O(nlogn) if limit >>n
Space complexity: O(limit) or O(n)