Leetcode 1020. Number of Enclaves

Input: grid1 = [[1,1,1,0,0],[0,1,1,1,1],[0,0,0,0,0],[1,0,0,0,0],[1,1,0,1,1]], grid2 = [[1,1,1,0,0],[0,0,1,1,1],[0,1,0,0,0],[1,0,1,1,0],[0,1,0,1,0]]
Output: 3
Explanation: In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
The 1s colored red in grid2 are those considered to be part of a sub-island. There are three sub-islands.
Input: grid1 = [[1,0,1,0,1],[1,1,1,1,1],[0,0,0,0,0],[1,1,1,1,1],[1,0,1,0,1]], grid2 = [[0,0,0,0,0],[1,1,1,1,1],[0,1,0,1,0],[0,1,0,1,0],[1,0,0,0,1]]
Output: 2
Explanation: In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
The 1s colored red in grid2 are those considered to be part of a sub-island. There are two sub-islands.
class Solution {
public:
int rsize{0};
int csize{0};
int count{0};
vector<vector<int>> next{{1,0}, {-1,0}, {0,1}, {0,-1}};

int dfs(vector<vector<int>>& grid1, vector<vector<int>>& grid2, vector<vector<bool>>& visited, int r, int c, bool& sea) {
if (r == rsize || r < 0 || c == csize || c < 0 || visited[r][c] || grid2[r][c] == 0) {
return 0;
}

if (grid1[r][c] == 0) {
sea = true;
}
visited[r][c] = true;
for (int i=0; i<4; i++) {
dfs(grid1, grid2, visited, r+next[i][0], c+next[i][1], sea);
}
if (sea)
return 0;

return 1;
}

int countSubIslands(vector<vector<int>>& grid1, vector<vector<int>>& grid2) {
rsize = grid1.size();
csize = grid1[0].size();
vector<vector<bool>> visited(rsize, vector<bool>(csize, false));
for (int i=0; i<rsize; i++) {
for (int j=0; j<csize; j++) {
if (grid1[i][j]) {
bool sea = false;
count += dfs(grid1, grid2, visited, i, j, sea);
}
}
}
return count;
}
};

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