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+// 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
+}