1674. Minimum Moves to Make Array Complementary
Description
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.
Solutions
Solution 1: Difference Array
Assume that in the final array, the sum of the pair $\textit{nums}[i]$ and $\textit{nums}[n-i-1]$ is $s$.
Let's denote $x$ as the smaller value between $\textit{nums}[i]$ and $\textit{nums}[n-i-1]$, and $y$ as the larger value.
For each pair of numbers, we have the following scenarios:
- If no replacement is needed, then $x + y = s$.
- If one replacement is made, then $x + 1 \le s \le y + \textit{limit}$.
- If two replacements are made, then $2 \le s \le x$ or $y + \textit{limit} + 1 \le s \le 2 \times \textit{limit}$.
That is:
- In the range $[2,..x]$, $2$ replacements are needed.
- In the range $[x+1,..x+y-1]$, $1$ replacement is needed.
- At $[x+y]$, no replacement is needed.
- In the range $[x+y+1,..y + \textit{limit}]$, $1$ replacement is needed.
- In the range $[y + \textit{limit} + 1,..2 \times \textit{limit}]$, $2$ replacements are needed.
We enumerate each pair of numbers and use a difference array to update the number of replacements needed in different ranges for each pair.
Finally, we find the minimum value among the prefix sums from index $2$ to $2 \times \textit{limit}$, which is the minimum number of replacements needed.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.
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