Solution 1

Cell \((i, j)\) has four edge-adjacent cells:

• cell \((i-1,\ j)\)
• cell \((i+1,\ j)\)
• cell \((i,\ j-1)\)
• cell \((i,\ j+1)\)

Inspect each of them to check if it exists.

Solution 2

If the size of the grid is \(1 \times 1\), the answer for cell \((1,1)\) is \(0\).

If the size of the grid is \(1 \times W\) (\(W > 1\)), the answers for cells \((1,1), (1,W)\) are \(1\), and the others are \(2\).

If the size of the grid is \(H \times 1\) (\(H > 1\)), the answers for cells \((1,1), (H,1)\) are \(1\), and the others are \(2\).

If the size of the grid is \(H \times W\) (\(H,W > 1\)), the answer is:

• \(2\) for cells \((1,1), (1,W), (H,1), (H,W)\);
• otherwise, \(3\) for the cells in at least one of row \(1\), row \(H\), column \(1\), or column \(W\);
• otherwise, \(4\).

Handle these cases correctly, and your solution will be accepted.

Sample code (C++, Solution 1)

#include <iostream>
using std::cin;
using std::cout;

int main (void) {
	int h, w;
	cin >> h >> w;

	int ans[h][w];
	for (int i = 0; i < h; i++) {
		for (int j = 0; j < w; j++) {
			ans[i][j] = 0;

			if (i-1 >= 0) ans[i][j]++;
			if (i+1 <  h) ans[i][j]++;
			if (j-1 >= 0) ans[i][j]++;
			if (j+1 <  w) ans[i][j]++;
		}
	}

	for (int i = 0; i < h; i++) {
		for (int j = 0; j < w; j++) {
			if (j > 0) { cout << " "; }
			cout << ans[i][j];
		}
		cout << "\n";
	}

	return 0;
}