Question
Formatted question description: https://leetcode.ca/all/1674.html
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 (0indexed), 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[n1i] = 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.
Algorithm
We think according to the piecewise function, assuming that a and b are a pair, that is, nums[i]=a, nums[ni1]=b.
 First, if nothing is changed, that is, 0 times, we get
a+b
.  If you change it once and make it bigger, then you should change a to make it as large as possible to get
limit+b
 if you change it to smaller, then you should change b to make it smaller, and the minimum is 1 to get
1+a
.  If you change it twice, if you change it to a larger value, you can at least get
limit+b+1
, that is, a becomes larger and b becomes 1 larger than itself;  similarly, if you change it to a smaller value, you can get the minimum of
2
.
So we got the key points:
 [2, a], need to change 2 times;
 [a+1, a+b1], need to change 1 time;
a+b
, need to change 0 times; [a+b+1, limit+b], need Change it once;
 finally [limit+b+1, 2limit] needs to be changed twice. Here we can think that 2limit is the maximum value because the limit given in constraints is always greater than or equal to nums[i].
Therefore, the important point is 2, a+1, a+b, a+b+1, limit+b+1, using the difference, assuming that the original initial number of changes is 0, starting from the above point, the number of changes is +2 (2), 1(1), 1(0), +1(1), +1(2), the actual number of changes to be changed in the brackets, thus forming a piecewise function. Open an array of size limit*2+2, and then for n/2 pairs of points, we calculate these points, and perform + oroperations on these points on the array. Calculate the prefix sum of this array and find the smallest moment, which is the answer.
Code
Java
Java

class Solution { public int minMoves(int[] nums, int limit) { int sumLimit = limit * 2; int[] differences = new int[sumLimit + 1]; for (int i = 0, j = nums.length  1; i < j; i++, j) { int sum = nums[i] + nums[j]; int curMinSum = Math.min(nums[i], nums[j]) + 1; int curMaxSum = Math.max(nums[i], nums[j]) + limit; differences[curMinSum]; differences[sum]; if (sum < sumLimit) differences[sum + 1]++; if (curMaxSum < sumLimit) differences[curMaxSum + 1]++; } int maxReduce = 0; for (int i = 1; i <= sumLimit; i++) { differences[i] += differences[i  1]; maxReduce = Math.min(maxReduce, differences[i]); } return nums.length + maxReduce; } }

// OJ: https://leetcode.com/problems/minimummovestomakearraycomplementary/ // Time: O(N + L) // Space: O(L) // Ref: https://leetcode.com/problems/minimummovestomakearraycomplementary/discuss/952773/PythonJavasimpleO(max(nk))method class Solution { public: int minMoves(vector<int>& A, int L) { vector<int> delta(2 * L + 2); // difference array int N = A.size(), ans = N, cur = 0; for (int i = 0; i < N / 2; ++i) { int a = A[i], b = A[N  i  1]; delta[2] += 2; delta[min(a, b) + 1]; delta[a + b]; delta[a + b + 1]++; delta[max(a, b) + L + 1]++; } for (int i = 2; i <= 2 * L; ++i) { cur += delta[i]; ans = min(ans, cur); } return ans; } };

print("Todo!")