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.

&nbsp;
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.


&nbsp;
Constraints:


	1 &lt;= original.length &lt;= 5 * 104
	1 &lt;= original[i] &lt;= 105
	1 &lt;= m, n &lt;= 4 * 104


class Solution:
    def construct2DArray(self, original: List[int], m: int, n: int) -> List[List[int]]:
        return [original[i:i+n] for i in range(0, len(original), n)] if m*n == len(original) else []
        
class Solution {
    public int[][] construct2DArray(int[] original, int m, int n) {        
        if (original.length != m * n) return new int[0][];
        
        int[][] ans = new int[m][n];
        int currRow = 0, currCol = 0;
        
        for (int num : original) {
            ans[currRow][currCol++] = num;
            
            if (currCol == n) {
                currCol = 0;
                currRow++;
            }
        }
        
        return ans;
    }
}

// TC: O(n), SC: O(m * n)
class Solution {
public:
    vector<vector<int>> construct2DArray(vector<int>& original, int m, int n) {
        if (m * n != original.size()) return {};

        vector<vector<int>> res;
        for (int i = 0; i < m*n; i+=n)
            res.push_back(vector<int>(original.begin()+i, original.begin()+i+n));

        return res;
    }
};
var construct2DArray = function(original, m, n) {
    if (original.length !== (m*n)) return []
    let result = []
    let arr = []
    for (let i = 0; i < original.length; i++){
        arr.push(original[i])
        if (arr.length === n){
            result.push(arr)
            arr = []
        }
    }
    return result
};
Convert 1D Array Into 2D Array