Introduction

In this blog post, we will explore how to solve the LeetCode problem "2216" 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 integer array nums. The array nums is beautiful if: nums.length is even. nums[i] != nums[i + 1] for all i % 2 == 0. Note that an empty array is considered beautiful. You can delete any number of elements from nums. When you delete an element, all the elements to the right of the deleted element will be shifted one unit to the left to fill the gap created and all the elements to the left of the deleted element will remain unchanged. Return the minimum number of elements to delete from nums to make it beautiful. Example 1: Input: nums = [1,1,2,3,5] Output: 1 Explanation: You can delete either nums[0] or nums[1] to make nums = [1,2,3,5] which is beautiful. It can be proven you need at least 1 deletion to make nums beautiful. Example 2: Input: nums = [1,1,2,2,3,3] Output: 2 Explanation: You can delete nums[0] and nums[5] to make nums = [1,2,2,3] which is beautiful. It can be proven you need at least 2 deletions to make nums beautiful. Constraints: 1 <= nums.length <= 105 0 <= nums[i] <= 105

Explanation

Here's an efficient approach to solve this problem:

• Greedy Approach: Iterate through the array and maintain a "beautiful" subarray. If a violation occurs (even index has the same value as the next element), delete the current element. Crucially, only increment the index when an element is kept.

• Even Length Maintenance: Because a beautiful array must have even length, if after processing the entire array, the length is odd, we need to delete one more element.

• Runtime Complexity: O(n), Storage Complexity: O(1)

Code

    def minDeletion(nums):
    """
    Calculates the minimum number of elements to delete to make the array beautiful.

    Args:
        nums: A list of integers.

    Returns:
        The minimum number of deletions required.
    """
    deletions = 0
    beautiful_length = 0
    i = 0
    while i < len(nums):
        if beautiful_length % 2 == 0:
            if i + 1 < len(nums) and nums[i] == nums[i + 1]:
                deletions += 1
                i += 1
            else:
                beautiful_length += 1
                i += 1
        else:
            beautiful_length += 1
            i += 1

    if beautiful_length % 2 != 0:
        deletions += 1

    return deletions