Introduction

In this blog post, we will explore how to solve the LeetCode problem "1674" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the breakdown of the solution:

• Difference Array/Prefix Sum: The core idea is to use a difference array (prefix sum) to efficiently track the number of moves required to achieve a specific sum for each pair (nums[i], nums[n-1-i]).
• Range Counts: We iterate through the pairs and update the difference array based on the possible scenarios: 0 moves (sum is already the target), 1 move (one number needs to be changed), or 2 moves (both numbers need to be changed).

• Minimize Moves: After processing all pairs, we compute the prefix sum of the difference array. The minimum value in this prefix sum array corresponds to the minimum number of moves needed to make the array complementary.

• Runtime Complexity: O(n), where n is the length of the input array nums.

• Storage Complexity: O(limit + n), where n is the length of the input array nums and limit is the integer limit.

Code

    def minMoves(nums, limit):
    n = len(nums)
    diff = [0] * (2 * limit + 2)

    for i in range(n // 2):
        a, b = nums[i], nums[n - 1 - i]
        diff[2] += 2
        diff[min(a, b) + 1] -= 1
        diff[a + b] -= 1
        diff[a + b + 1] += 1
        diff[max(a, b) + limit + 1] += 1
        diff[2 * limit + 1] -= 2

    moves = 2 * n
    curr = 0
    for i in range(2, 2 * limit + 1):
        curr += diff[i]
        moves = min(moves, curr)

    return moves