Leetcode 2022 Convert 1-D array to 2-D array

Problem Statement — 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

Time: O(m * n)
Space : O(m * n)

class Solution {
public:
    vector<vector<int>> construct2DArray(vector<int>& original, int m, int n) {
        if (original.size() != m*n) 
            return {};
        vector<vector<int>> ans(m, vector<int>(n,0));
        int k = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans[i][j] = original[k++];
            }
        }
        return ans;
    }
};

Time: O(m * n)
Space : O(m * n)

class Solution {
public:
    vector<vector<int>> construct2DArray(vector<int>& original, int m, int n) {
        if (original.size() != m*n) 
            return {};
        vector<vector<int>> ans(m, vector<int>(n,0));
        for(int i=0;i<original.size();i++)
		    ans[i/n][i%n] = original[i];
        return ans;
    }
};

Feel free to ask any question in the comment section.
I hope that you’ve found the solution useful.