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-rw-r--r--num_enclaves.cpp45
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diff --git a/num_enclaves.cpp b/num_enclaves.cpp
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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
-}