Rotten oranges

class Solution
{
    public:
    //Function to find minimum time required to rot all oranges.
    int orangesRotting(vector<vector<int>>& grid) {
        deque<pair<int,int>> q;
        int row=grid.size(),col=grid[0].size();
        vector<vector<bool>> check(row,vector<bool>(col,false));
        int total_fresh=0,ans=0;
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                if(grid[i][j]==2){ q.push_back(make_pair(i,j)); check[i][j]=true;}
                if(grid[i][j]==1) total_fresh++;
            }
        }
        while(q.size() && total_fresh){
            int k=q.size();
            int x[4]={-1,0,1,0};
            int y[4]={0,1,0,-1};
            for(int i=0;i<k;i++){
                pair<int,int> temp=q.front();
                q.pop_front();
                for(int k=0;k<4;k++){
                    int new_x=temp.first+x[k];
                    int new_y=temp.second+y[k];
                    if(new_x>=0 && new_y>=0 && new_x<row && new_y<col && grid[new_x][new_y]==1 && !check[new_x][new_y]){
                        q.push_back(make_pair(new_x,new_y));
                        check[new_x][new_y]=true;
                        total_fresh--;
                    }
                }
            }
            ans++;
        }
        if(total_fresh!=0) return -1;
        return ans;
    }
};