Surrounded regions

/*
Logic:This is proper DFS problem.
observation:Only that bulk of O's which is on the corner will be remained O.
so we run dfs for every corner zero and change it to small o that means it will remain
We are changing it into small o so that we can differentiate between which O's to be
O and which to be changed.
O after all operations.
rest all O will be captured.
*/

void dfs(vector<vector<char>> &board,int i,int j,int r,int c){
    int row[4]={-1,0,1,0};
    int col[4]={0,1,0,-1};
    board[i][j]='o';
    for(int k=0;k<4;k++){
         int new_i=i+row[k];
         int new_j=j+col[k];
         if(new_i>=0 && new_i<r && new_j>=0 && new_j<c && board[new_i][new_j]=='O') dfs(board,new_i,new_j,r,c);
    }
}

class Solution {
public:
    void solve(vector<vector<char>>& board) {
        int r=board.size();
        int c=board[0].size();
        for(int i=0;i<r;i++){
            for(int j=0;j<c;j++){
                if((i==0 || i==r-1 || j==0 || j==c-1 ) && board[i][j]=='O')dfs(board,i,j,r,c);
            }
        }

        for(int i=0;i<r;i++){
            for(int j=0;j<c;j++){
                if(board[i][j]=='o') board[i][j]='O';
                else board[i][j]='X';
            }
        }
    }
};