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: Brute Force

the i-th element in original array will have index (i//n, i % n) in the 2D array.

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

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));
    for (int i = 0; i < m * n; ++i)
      ans[i / n][i % n] = original[i];
    return ans;
  }
};

The post 花花酱 LeetCode 2022. Convert 1D Array Into 2D Array appeared first on Huahua's Tech Road.

If you don’t like grid[y][x], you can use grid[r][c] instead.

#include <iostream>

using namespace std;

int main(int argc, char** argv) {
  constexpr int kRows = 2;
  constexpr int kCols = 3;
  int arr[kRows][kCols] = {{1,2,3},{4,5,6}};
  for (int r = 0; r < kRows; ++r) {
    for (int c = 0; c < kCols; ++c)
      cout << arr[r][c] << " ";
    cout << endl;
  }

  for (int r = 0; r < kRows; ++r)
    for (int c = 0; c < kCols; ++c)
      cout << &(arr[r][c]) - &(arr[0][0]) << " ";
  return 0;
}

// Output:
// 1 2 3
// 4 5 6
// 0 1 2 3 4 5

The post Why grid[y][x] instead of grid[x][y]? appeared first on Huahua's Tech Road.