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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[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.
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
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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] += 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] += 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 = [0] * (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] += 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] += 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 }