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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 * 10<sup>4</sup>1 <= original[i] <= 10<sup>5</sup>1 <= m, n <= 4 * 10<sup>4</sup>
Solution
So in the question we just have to divide the single array into a 2 dimensional Matrix which is M X N and original is our array.
How do we solve it?
class Solution:
def construct2DArray(self, original: List[int], m: int, n: int) -> List[List[int]]:
tmp=[]
res=[]
if len(original)!=m*n: # eg. 2*2=4 and if og is 1 then this will be handled
return []
for i in range(len(original)):
tmp.append(original[i])
if (i+1)%n==0:
res.append(tmp)
tmp=[]
return res