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
SOLUTION:
class Solution:
def construct2DArray(self, original: List[int], m: int, n: int) -> List[List[int]]:
k = len(original)
if k != m * n:
return []
for i in range(m):
original.extend([original[i * n : (i + 1) * n]])
return original[k:]
Top comments (0)