#
# Create by Hua on 2/1/22.
#

"""
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 * 10^4
    1 <= original[i] <= 10^5
    1 <= m, n <= 4 * 10^4

"""


class Solution(object):
    def construct2DArray(self, original, m, n):
        """
        :type original: List[int]
        :type m: int
        :type n: int
        :rtype: List[List[int]]


        easy 5 min, need to work out the math.
        thought: is it possible if m*n < len(original)?
        hint2: It is possible if and only if m * n == original.length

        [0, n-1]
        [n, 2*n -1]
        ...

        02/01/2022 14:18	Accepted	1026 ms	21.9 MB	python
        """
        ret = list()
        size = len(original)
        if m*n != size : return ret

        for i in range(m):
            ls = original[i*n:(i+1)*n]  # key of this problem.
            ret.append(ls)

        return ret





