Introduction

In this blog post, we will explore how to solve the LeetCode problem "2610" 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. You need to create a 2D array from nums satisfying the following conditions: The 2D array should contain only the elements of the array nums. Each row in the 2D array contains distinct integers. The number of rows in the 2D array should be minimal. Return the resulting array. If there are multiple answers, return any of them. Note that the 2D array can have a different number of elements on each row. Example 1: Input: nums = [1,3,4,1,2,3,1] Output: [[1,3,4,2],[1,3],[1]] Explanation: We can create a 2D array that contains the following rows: - 1,3,4,2 - 1,3 - 1 All elements of nums were used, and each row of the 2D array contains distinct integers, so it is a valid answer. It can be shown that we cannot have less than 3 rows in a valid array. Example 2: Input: nums = [1,2,3,4] Output: [[4,3,2,1]] Explanation: All elements of the array are distinct, so we can keep all of them in the first row of the 2D array. Constraints: 1 <= nums.length <= 200 1 <= nums[i] <= nums.length

Explanation

Here's a breakdown of the solution:

• Frequency Counting: Determine the frequency of each number in the input array nums.
• Iterative Row Creation: Build the 2D array row by row. For each row, iterate through the number frequencies. If a number's frequency is greater than zero, add it to the current row and decrement its frequency.

• Termination: Continue creating rows until all numbers have been added to the 2D array (i.e., all frequencies are zero).

• Runtime Complexity: O(N*M) where N is the length of the input array and M is the maximum frequency of any number in the array. In the worst-case scenario, M can be equal to N which will make it O(N^2).

• Storage Complexity: O(N), due to the frequency map and the resulting 2D array.

Code

    def create_2d_array(nums):
    """
    Creates a 2D array from the input array nums, satisfying the given conditions.

    Args:
        nums: A list of integers.

    Returns:
        A list of lists of integers, representing the resulting 2D array.
    """

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

    result = []
    while any(counts.values()):
        row = []
        for num in list(counts.keys()):  # Iterate over a copy to allow deletion
            if counts[num] > 0:
                row.append(num)
                counts[num] -= 1
                if counts[num] == 0:
                    del counts[num]
        result.append(row)

    return result