class Solution {
    public:
    vector<int> dx{0, 1, -1, 0},dy{1, 0, 0, -1};
    bool isVisitable(int x, int y, int N, int M){
        return x>=0 && y>=0 && x<N && y<M;
    }
    void dfs(int i, int j, int N, int M, vector<vector<int>>& matrix){
        for(int k =0 ; k <4; k++){
            int x = i + dx[k], y = j + dy[k];
            if( isVisitable(x, y, N, M) && matrix[x][y] != 0)
                matrix[x][y]=0,dfs(x, y, N, M, matrix);
        }
    }
    int closedIslands(vector<vector<int>>& matrix, int N, int M) {
        // Code here
        for(int i = 0; i < N; i++){
            for(int j = 0; j < M ; j++){
                if( (i == 0 || j == 0 || i == N -1 || j == M - 1 )&& matrix[i][j] == 1){
                    //check if 1 exists on the edges of matrix
                    //if yes visit all the cells that have 1 and are 
                    //reachable from current cell and mark them visited by making 0
                    matrix[i][j] = 0;
                    dfs(i, j, N, M,matrix);
                }
            }
        }
        
        int ans = 0;
        for(int i = 0 ; i < N; i++){
            for(int j = 0; j < M ; j++){
                if(matrix[i][j] == 1){
                    //whatever cells not visited in previous case 
                    //we visit now and mark visited(0) and increment the 
                    //the answer
                    matrix[i][j] = 0;
                    dfs(i, j ,N, M, matrix);
                    ans++;
                }
            }
        }
        return ans;
    }
};