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)

// Author: Huahua
class Solution {
public:
  int minMoves(vector<int>& nums, int limit) {
    const int n = nums.size();
    vector<int> delta(limit * 2 + 2);
    for (int i = 0; i < n / 2; ++i) {
      const int a = min(nums[i], nums[n - i - 1]);
      const int b = max(nums[i], nums[n - i - 1]);
      delta[2] += 2;          // dec a, dec b
      --delta[a + 1];         // dec a 
      --delta[a + b];         // no op
      ++delta[a + b + 1];     // inc a
      ++delta[b + limit + 1]; // inc a, inc b     
    }
    int ans = n;
    for (int t = 2, cur = 0; t <= limit * 2; ++t) {
      cur += delta[t];
      ans = min(ans, cur);
    }
    return ans;
  }
};

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