# Solving Leetcode Interviews in Seconds with AI: Split the Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3046" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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. You have to split the array into two parts nums1 and nums2 such that:  nums1.length == nums2.length == nums.length / 2. nums1 should contain distinct elements. nums2 should also contain distinct elements.  Return true if it is possible to split the array, and false otherwise.   Example 1:  Input: nums = [1,1,2,2,3,4] Output: true Explanation: One of the possible ways to split nums is nums1 = [1,2,3] and nums2 = [1,2,4].  Example 2:  Input: nums = [1,1,1,1] Output: false Explanation: The only possible way to split nums is nums1 = [1,1] and nums2 = [1,1]. Both nums1 and nums2 do not contain distinct elements. Therefore, we return false.    Constraints:  1 <= nums.length <= 100 nums.length % 2 == 0  1 <= nums[i] <= 100  

	# Explanation
	Here's a breakdown of the approach, complexities, and the Python code:

*   **High-Level Approach:**
    *   Count the frequency of each number in the input array.
    *   Calculate the number of unique elements.
    *   Check if the number of duplicates (total elements - unique elements) is less than or equal to half the length of the array. If it is, a split is possible.

*   **Complexity:**
    *   Runtime: O(n), where n is the length of the input array.
    *   Storage: O(k), where k is the range of numbers in the input array, which is capped at 100 in the constraints, hence practically O(1).

	
	# Code
	```python
	def can_split_array(nums):
    """
    Checks if the array can be split into two arrays with distinct elements.

    Args:
        nums: An integer array of even length.

    Returns:
        True if the array can be split, False otherwise.
    """

    n = len(nums)
    counts = {}
    for num in nums:
        counts[num] = counts.get(num, 0) + 1

    unique_count = len(counts)
    duplicate_count = n - unique_count

    return duplicate_count <= n // 2
	```