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Diffstat (limited to 'other/num_enclaves.cpp')
-rw-r--r-- | other/num_enclaves.cpp | 45 |
1 files changed, 45 insertions, 0 deletions
diff --git a/other/num_enclaves.cpp b/other/num_enclaves.cpp new file mode 100644 index 0000000..2629d65 --- /dev/null +++ b/other/num_enclaves.cpp @@ -0,0 +1,45 @@ +// 1020. Number of Enclaves +#include "leetcode.h" + +/* + * given an 'm * n' binary matrix grid, where 0 represents a sea + * cell and 1 represents a land cell. a move consists of walking + * from one land cell to another adjacent (4-directionally) land + * cell o walking off the boundary of the 'grid'. return number + * of land cells in 'grid' for which we cannot walk off the + * boundary of the grid in any number of moves. + */ + +class Solution { + int dfs(vvd & grid, int i, int j) { + if (i < 0 || j < 0 || i >= grid.size() || j >= grid[0].size() || + grid[i][j] == 0) + return 0; + grid[i][j] = 0; + return 1 + dfs(grid, i - 1, j) + dfs(grid, i + 1, j) + dfs(grid, i, j - 1) + + dfs(grid, i, j + 1); + } + +public: + int numEnclaves(vvd & grid) { + int ones = 0, zeros = 0; + for (int i = 0; i < grid.size(); i++) + for (int j = 0; j < grid[0].size(); j++) + ones += grid[i][j]; + for (int i = 0; i < grid.size(); i++) + for (int j = 0; j < grid[0].size(); j++) + if ((i == 0 || j == 0 || i == grid.size() - 1 || + j == grid[0].size() - 1) && + grid[i][j] == 1) + zeros += dfs(grid, i, j); + return ones - zeros; + } +}; + +int main() { + Solution obj; + vvd g1 = {{0, 0, 0, 0}, {1, 0, 1, 0}, {0, 1, 1, 0}, {0, 0, 0, 0}}; + vvd g2 = {{0, 1, 1, 0}, {0, 0, 1, 0}, {0, 0, 1, 0}, {0, 0, 0, 0}}; + printf("%d\n", obj.numEnclaves(g1)); // expect: 3 + printf("%d\n", obj.numEnclaves(g2)); // expect: 0 +} |