convert-1d-array-into-2d-array.cpp

//question//


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.
Example 4:

Input: original = [3], m = 1, n = 2
Output: []
Explanation:
There is 1 element in original.
It is impossible to make 1 element fill all the spots in a 1x2 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 {
public:
    vector<vector<int>> construct2DArray(vector<int>& original, int m, int n) {
        vector<vector<int>> v;
        int k=0;
        if(original.size()!=m*n)
        {
            return v;
        }
        for(int i=0;i<m;i++)
        {
            vector<int> v1;
            for(int j=0;j<n;j++)
            {
                v1.push_back(original[k]);
                k++;
            }
            v.push_back(v1);
        }
        return v;
    }
};