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 numscomplementary.
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.
class Solution {
public int minMoves(int[] nums, int limit) {
/*
For each pair of numbers (at index i and N - 1 - i) l and r:
By 2 moves we can get any number between [2, limit*2] (including move a number to the same number)
After only one move (change one of the numbers to a number between 1 and limit)
The minimum sum we can get is (min(l, r) + 1) (let this be oneMoveMin)
The maximum sum we can get is (max(l, r) + limit) (let this be oneMoveMax)
We need no move to get (l + r) (let this be justGood)
Therefore, to get:
[~, oneMoveMin - 1] - 2 moves
[oneMoveMin, justGood-1] - 1 move
[justGood] - 0 move
[justGood + 1, oneMoveMax] - 1 move
[oneMoveMax + 1, ~] - 2 moves
*/
int[] moves = new int[2 * limit + 2];
int left = 0, right = nums.length - 1;
while (left < right) {
int l = nums[left ++];
int r = nums[right --];
int oneMoveMin = Math.min(l, r) + 1;
int oneMoveMax = Math.max(l, r) + limit;
int justGood = l + r;
moves[2] += 2;
moves[oneMoveMin] -= 1;
moves[justGood] -= 1;
moves[justGood + 1] += 1;
moves[oneMoveMax + 1] += 1;
}
int res = nums.length / 2, curr = 0;
for (int i = 2; i <= 2 * limit; i ++) {
curr += moves[i];
res = Math.min(res, curr);
}
return res;
}
}