Introduction

In this blog post, we will explore how to solve the LeetCode problem "2170" 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 a 0-indexed array nums consisting of n positive integers. The array nums is called alternating if: nums[i - 2] == nums[i], where 2 <= i <= n - 1. nums[i - 1] != nums[i], where 1 <= i <= n - 1. In one operation, you can choose an index i and change nums[i] into any positive integer. Return the minimum number of operations required to make the array alternating. Example 1: Input: nums = [3,1,3,2,4,3] Output: 3 Explanation: One way to make the array alternating is by converting it to [3,1,3,1,3,1]. The number of operations required in this case is 3. It can be proven that it is not possible to make the array alternating in less than 3 operations. Example 2: Input: nums = [1,2,2,2,2] Output: 2 Explanation: One way to make the array alternating is by converting it to [1,2,1,2,1]. The number of operations required in this case is 2. Note that the array cannot be converted to [2,2,2,2,2] because in this case nums[0] == nums[1] which violates the conditions of an alternating array. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 105

Explanation

Here's the solution:

• Frequency Analysis: Count the frequency of each number at even and odd indices separately.
• Optimal Choices: Find the two most frequent numbers for both even and odd indices. The goal is to maximize the usage of frequent numbers to minimize changes.

• Conflict Resolution: If the most frequent numbers in even and odd indices are the same, consider using the second most frequent numbers to avoid conflicts.

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

• Space Complexity: O(n), for frequency counting.

Code

    from collections import Counter

def minimumOperations(nums):
    n = len(nums)
    even_nums = nums[::2]
    odd_nums = nums[1::2]

    even_counts = Counter(even_nums)
    odd_counts = Counter(odd_nums)

    even_most_common = even_counts.most_common(2)
    odd_most_common = odd_counts.most_common(2)

    even_first = even_most_common[0][0]
    even_first_count = even_most_common[0][1]
    even_second = even_most_common[1][0] if len(even_most_common) > 1 else None
    even_second_count = even_most_common[1][1] if len(even_most_common) > 1 else 0

    odd_first = odd_most_common[0][0]
    odd_first_count = odd_most_common[0][1]
    odd_second = odd_most_common[1][0] if len(odd_most_common) > 1 else None
    odd_second_count = odd_most_common[1][1] if len(odd_most_common) > 1 else 0

    even_len = len(even_nums)
    odd_len = len(odd_nums)

    if even_first != odd_first:
        return even_len - even_first_count + odd_len - odd_first_count
    else:
        option1 = even_len - even_first_count + odd_len - odd_second_count if odd_second is not None else float('inf')
        option2 = even_len - even_second_count + odd_len - odd_first_count if even_second is not None else float('inf')
        return min(option1, option2)