Introduction

In this blog post, we will explore how to solve the LeetCode problem "3224" 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 size n where n is even, and an integer k. You can perform some changes on the array, where in one change you can replace any element in the array with any integer in the range from 0 to k. You need to perform some changes (possibly none) such that the final array satisfies the following condition: There exists an integer X such that abs(a[i] - a[n - i - 1]) = X for all (0 <= i < n). Return the minimum number of changes required to satisfy the above condition. Example 1: Input: nums = [1,0,1,2,4,3], k = 4 Output: 2 Explanation: We can perform the following changes: Replace nums[1] by 2. The resulting array is nums = [1,2,1,2,4,3]. Replace nums[3] by 3. The resulting array is nums = [1,2,1,3,4,3]. The integer X will be 2. Example 2: Input: nums = [0,1,2,3,3,6,5,4], k = 6 Output: 2 Explanation: We can perform the following operations: Replace nums[3] by 0. The resulting array is nums = [0,1,2,0,3,6,5,4]. Replace nums[4] by 4. The resulting array is nums = [0,1,2,0,4,6,5,4]. The integer X will be 4. Constraints: 2 <= n == nums.length <= 105 n is even. 0 <= nums[i] <= k <= 105

Explanation

Here's the approach to solve this problem:

• Iterate and Group: Iterate through the first half of the nums array. For each i, consider the pair (nums[i], nums[n - i - 1]). These pairs must satisfy the condition abs(a[i] - a[n - i - 1]) = X for some X.
• Analyze Pair Differences: For each pair, calculate the absolute difference and store it. Also, keep track of the frequency of each number within the pair.

• Minimize Changes: To minimize the number of changes, for each pair, we have two options: make both numbers equal to a new number, or keep the initial difference and modify elements accordingly. We analyze frequencies and differences to determine the minimal number of changes for that pair.

• Runtime & Storage Complexity: O(n + k), O(k) where n is the length of the array nums and k is the largest number in the array, this comes from the constraint 0 <= nums[i] <= k <= 10^5, also the creation of the count array of size k.

Code

    def min_changes(nums, k):
    n = len(nums)
    changes = 0

    for i in range(n // 2):
        a = nums[i]
        b = nums[n - i - 1]

        if a != b:
            changes += 1

    diff_counts = {}
    num_counts = {}

    for i in range(n // 2):
        a = nums[i]
        b = nums[n - i - 1]

        diff = abs(a - b)

        if diff not in diff_counts:
            diff_counts[diff] = 0
        diff_counts[diff] += 1

        if a not in num_counts:
            num_counts[a] = 0
        num_counts[a] += 1

        if b not in num_counts:
            num_counts[b] = 0
        num_counts[b] += 1

    total_pairs = n // 2
    min_changes_needed = float('inf')

    for diff in diff_counts:
        changes_needed = total_pairs - diff_counts[diff]

        min_changes_needed = min(min_changes_needed, changes_needed)

    num_freq = [0] * (k + 1)
    for i in range(n//2):
        num_freq[nums[i]] += 1
        num_freq[nums[n-1-i]] += 1

    max_freq = 0
    for i in range(k+1):
        max_freq = max(max_freq, num_freq[i])

    min_changes_needed = min(min_changes_needed, total_pairs * 2 - max_freq)

    num_unique_val_pairs = 0
    for i in range(n//2):
        if nums[i] != nums[n-i-1]:
            num_unique_val_pairs += 1

    counts = {}
    for i in range(n//2):
        diff = abs(nums[i] - nums[n-i-1])
        if diff not in counts:
            counts[diff] = 0
        counts[diff] += 1

    most_frequent_diff = 0
    for key in counts:
        most_frequent_diff = max(most_frequent_diff, counts[key])

    min_changes_needed = min(min_changes_needed, num_unique_val_pairs - most_frequent_diff + (n//2 - num_unique_val_pairs) * 2)

    counts = {}
    for i in range(n//2):
      a = nums[i]
      b = nums[n-i-1]
      if a==b: continue
      if abs(a-b) not in counts:
        counts[abs(a-b)] = 0
      counts[abs(a-b)] +=1

    most_freq = 0
    for val in counts:
      most_freq = max(counts[val], most_freq)

    no_change_pairs = n//2 - num_unique_val_pairs

    if num_unique_val_pairs == 0:
      return 0

    min_changes_needed = min(min_changes_needed, num_unique_val_pairs - most_freq + no_change_pairs*2)

    return min_changes_needed