Type an input and write what you think is the output . Each problem is converted to Web Assembly, so any possible input will show the corresponding output. Only correct predictions count.
Locked until you have predicted 3 outputs correctly.
| 1 | vector<vector<int>> solution(vector<vector<int>> grid) { |
| 2 | |
| 3 | if (grid.empty() || grid[0].empty()) { |
| 4 | return grid; |
| 5 | } |
| 6 | |
| 7 | int rows = grid.size(); |
| 8 | int cols = grid[0].size(); |
| 9 | |
| 10 | queue<pair<int, int>> edge; |
| 11 | |
| 12 | for (int r = 0; r < rows; r++) { |
| 13 | for (int c = 0; c < cols; c++) { |
| 14 | |
| 15 | if (grid[r][c] == 0) { |
| 16 | edge.push({r, c}); |
| 17 | } |
| 18 | } |
| 19 | } |
| 20 | |
| 21 | const int dr[] = {1, -1, 0, 0}; |
| 22 | const int dc[] = {0, 0, 1, -1}; |
| 23 | |
| 24 | while (!edge.empty()) { |
| 25 | |
| 26 | auto [r, c] = edge.front(); |
| 27 | edge.pop(); |
| 28 | |
| 29 | for (int d = 0; d < 4; d++) { |
| 30 | |
| 31 | int nr = r + dr[d]; |
| 32 | int nc = c + dc[d]; |
| 33 | |
| 34 | if (nr < 0 || nr >= rows || nc < 0 || nc >= cols) { |
| 35 | continue; |
| 36 | } |
| 37 | |
| 38 | if (grid[nr][nc] != INT_MAX) { |
| 39 | continue; |
| 40 | } |
| 41 | |
| 42 | grid[nr][nc] = grid[r][c] + 1; |
| 43 | edge.push({nr, nc}); |
| 44 | } |
| 45 | } |
| 46 | |
| 47 | return grid; |
| 48 | } |
Names have been stripped. The signature is the only clue you get for free. Compiled as C++20 with the standard headers and using namespace std; already in scope.