# Solving Leetcode Interviews in Seconds with AI: Convert 1D Array Into 2D Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2022" 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 a 0-indexed 1-dimensional (1D) integer array original, and two integers, m and n. You are tasked with creating a 2-dimensional (2D) array with  m rows and n columns using all the elements from original. The elements from indices 0 to n - 1 (inclusive) of original should form the first row of the constructed 2D array, the elements from indices n to 2 * n - 1 (inclusive) should form the second row of the constructed 2D array, and so on. Return an m x n 2D array constructed according to the above procedure, or an empty 2D array if it is impossible.   Example 1:   Input: original = [1,2,3,4], m = 2, n = 2 Output: [[1,2],[3,4]] Explanation: The constructed 2D array should contain 2 rows and 2 columns. The first group of n=2 elements in original, [1,2], becomes the first row in the constructed 2D array. The second group of n=2 elements in original, [3,4], becomes the second row in the constructed 2D array.  Example 2:  Input: original = [1,2,3], m = 1, n = 3 Output: [[1,2,3]] Explanation: The constructed 2D array should contain 1 row and 3 columns. Put all three elements in original into the first row of the constructed 2D array.  Example 3:  Input: original = [1,2], m = 1, n = 1 Output: [] Explanation: There are 2 elements in original. It is impossible to fit 2 elements in a 1x1 2D array, so return an empty 2D array.    Constraints:  1 <= original.length <= 5 * 104 1 <= original[i] <= 105 1 <= m, n <= 4 * 104  

	# Explanation
	Here's an efficient solution to the problem:

*   **Check feasibility:** Verify if the product of `m` and `n` equals the length of the `original` array. If not, return an empty list.
*   **Reshape:** Create the 2D array and populate it by iterating through the `original` array and placing elements into the correct row and column based on the row and column indices.
*   **Return result:** Return the new 2D array.

*   **Runtime Complexity:** O(m\*n), where m is the number of rows and n is the number of columns.
*   **Storage Complexity:** O(m\*n) to store the reshaped 2D array.

	
	# Code
	```python
	def construct2DArray(original, m, n):
    """
    Constructs a 2D array from a 1D array.

    Args:
        original (list): The 1D integer array.
        m (int): The number of rows in the 2D array.
        n (int): The number of columns in the 2D array.

    Returns:
        list: The constructed 2D array, or an empty list if impossible.
    """
    if m * n != len(original):
        return []

    result = [[0] * n for _ in range(m)]
    for i in range(m):
        for j in range(n):
            result[i][j] = original[i * n + j]

    return result
	```