Can you solve this real interview question? Convert 1D Array Into 2D Array - 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:
[https://assets.leetcode.com/uploads/2021/08/26/image-20210826114243-1.png]
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.
class Solution:
def construct2DArray(self, original: List[int], m: int, n: int) -> List[List[int]]:
# Check if the total elements match m * n
if n * m != len(original):
return []
ans = []
temp = []
cnt = 0
for val in original:
temp.append(val) # Add the current element to the temp row
cnt += 1
if cnt == n: # If temp has enough elements for one row
ans.append(temp) # Append the row to the 2D array
temp = [] # Reset temp for the next row
cnt = 0 # Reset the counter
return ans