# 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 (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 + nums = 1 + 3 = 4.
nums + nums = 2 + 2 = 4.
nums + nums = 2 + 2 = 4.
nums + nums = 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

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[n-i-1]=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+b-1], 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 + or-operations on these points on the array. Calculate the prefix sum of this array and find the smallest moment, which is the answer.

# Code

• 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;
}
}

############

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

d += 2;
d[limit * 2 + 1] -= 2;

d[a + 1] -= 1;
d[b + limit + 1] += 1;

d[a + b] -= 1;
d[a + b + 1] += 1;
}
int ans = n, s = 0;
for (int i = 2; i <= limit * 2; ++i) {
s += d[i];
if (ans > s) {
ans = s;
}
}
return ans;
}
}

• // OJ: https://leetcode.com/problems/minimum-moves-to-make-array-complementary/
// Time: O(N + L)
// Space: O(L)
// Ref: https://leetcode.com/problems/minimum-moves-to-make-array-complementary/discuss/952773/PythonJava-simple-O(max(n-k))-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;
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;
}
};

• class Solution:
def minMoves(self, nums: List[int], limit: int) -> int:
d =  * (limit * 2 + 2)
n = len(nums)

for i in range(n >> 1):
a, b = min(nums[i], nums[n - i - 1]), max(nums[i], nums[n - i - 1])

d += 2
d[limit * 2 + 1] -= 2

d[a + 1] -= 1
d[b + limit + 1] += 1

d[a + b] -= 1
d[a + b + 1] += 1

ans, s = n, 0
for v in d[2 : limit * 2 + 1]:
s += v
if ans > s:
ans = s
return ans


• func minMoves(nums []int, limit int) int {
d := make([]int, limit*2+2)
n := len(nums)
for i := 0; i < n>>1; i++ {
a, b := min(nums[i], nums[n-i-1]), max(nums[i], nums[n-i-1])
d += 2
d[limit*2+1] -= 2

d[a+1] -= 1
d[b+limit+1] += 1

d[a+b] -= 1
d[a+b+1] += 1
}
ans, s := n, 0
for _, v := range d[2 : limit*2+1] {
s += v
if ans > s {
ans = s
}
}
return ans
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

func min(a, b int) int {
if a < b {
return a
}
return b
}