Algorithm:

To implement this question we use BFS and a queue data structure.

1. At first, we push all positions into the queue which has 2 and make a partition by inserting NULL into the queue.
2. We pop every element from the queue until the first NULL comes and Go for its four neighbor's if there is any 1 then make it two and push it into the queue and separate this section again inserting a NULL into the queue.
3. Whenever we encounter NULL except for the first NULL and if it is not a last element of the queue then we increase the count value.
4. Repeat step 2 to 3 until the queue is empty.

C++ Implementation for Rotten Oranges problem

#include <iostream>
#include <bits/stdc++.h>
using namespace std;

bool zero(int *arr,int r,int c){
	for(int i=0;i<r;i++){
		for(int j=0;j<c;j++){
			if(*((arr+i*c)+j)==1)
				return false;
		}
	}
	return true;
} 

int rotten(int *arr,int r,int c){
	queue<pair<int,int> >q;
	int store=0,temp=0;
	for(int i=0;i<r;i++){
		for(int j=0;j<c;j++){
			if(*((arr+i*c)+j)==2){
				q.push(make_pair(i,j));
			}
		}
	}
	q.push(make_pair(-1,-1));
	while(!q.empty()){
		pair<int,int> p=q.front();
		q.pop();
		if(p.first!=-1){
			if(*((arr+p.first*c)+p.second)==1){
				temp=1;
				*((arr+p.first*c)+p.second)=2;
			}
			if(p.first==0 && p.second==0){
				if(*((arr+(p.first+1)*c)+p.second)==1){
					q.push(make_pair((p.first+1),p.second));
				}
				if(*((arr+(p.first)*c)+p.second+1)==1){
					q.push(make_pair((p.first),p.second+1));
				}
			}
			else if(p.first==0 && p.second==c-1){
				if(*((arr+(p.first+1)*c)+p.second)==1){
					q.push(make_pair((p.first+1),p.second));
				}
				if(*((arr+(p.first)*c)+p.second-1)==1){
					q.push(make_pair((p.first),p.second-1));
				}
			}
			else if(p.first==0){
				if(*((arr+(p.first+1)*c)+p.second)==1){
					q.push(make_pair((p.first+1),p.second));
				}
				if(*((arr+(p.first)*c)+p.second-1)==1){
					q.push(make_pair((p.first),p.second-1));
				}
				if(*((arr+(p.first)*c)+p.second+1)==1){
					q.push(make_pair((p.first),p.second+1));
				}
			}
			else if(p.first==r-1 && p.second==0){
				if(*((arr+(p.first-1)*c)+p.second)==1){
					q.push(make_pair((p.first-1),p.second));
				}
				if(*((arr+(p.first)*c)+p.second+1)==1){
					q.push(make_pair((p.first),p.second+1));
				}
			}
			else if(p.second==0){
				if(*((arr+(p.first-1)*c)+p.second)==1){
					q.push(make_pair((p.first-1),p.second));
				}
				if(*((arr+(p.first+1)*c)+p.second)==1){
					q.push(make_pair((p.first+1),p.second));
				}
				if(*((arr+(p.first)*c)+p.second+1)==1){
					q.push(make_pair((p.first),p.second+1));
				}
			}
			else if(p.first==r-1 && p.second==c-1){
				if(*((arr+(p.first-1)*c)+p.second)==1){
					q.push(make_pair((p.first-1),p.second));
				}
				if(*((arr+(p.first)*c)+p.second-1)==1){
					q.push(make_pair((p.first),p.second-1));
				}
			}
			else if(p.first==r-1){
				if(*((arr+(p.first-1)*c)+p.second)==1){
					q.push(make_pair((p.first-1),p.second));
				}
				if(*((arr+(p.first)*c)+p.second-1)==1){
					q.push(make_pair((p.first),p.second-1));
				}
				if(*((arr+(p.first)*c)+p.second+1)==1){
					q.push(make_pair((p.first),p.second+1));
				}
			}
			else if(p.second==c-1){
				if(*((arr+(p.first-1)*c)+p.second)==1){
					q.push(make_pair((p.first-1),p.second));
				}
				if(*((arr+(p.first+1)*c)+p.second)==1){
					q.push(make_pair((p.first+1),p.second));
				}
				if(*((arr+(p.first)*c)+p.second-1)==1){
					q.push(make_pair((p.first),p.second-1));
				}
			}
			else{
				if(*((arr+(p.first)*c)+p.second-1)==1){
					q.push(make_pair((p.first),p.second-1));
				}
				if(*((arr+(p.first)*c)+p.second+1)==1){
					q.push(make_pair((p.first),p.second+1));
				}
				if(*((arr+(p.first-1)*c)+p.second)==1){
					q.push(make_pair((p.first-1),p.second));
				}
				if(*((arr+(p.first+1)*c)+p.second)==1){
					q.push(make_pair((p.first+1),p.second));
				}
			}
		}
		if(p.first==-1){
			if(!q.empty()){
				q.push(make_pair(-1,-1));
			}
			if(temp==1){
				store++;
				temp=0;
			}
		}
	}
	return store;
}

int main() {
	int num;
	cin>>num;
	
	for(int i=0;i<num;i++){
		int r,c;
		cin>>r>>c;
		int arr[r][c];
		for(int j=0;j<r;j++){
			for(int k=0;k<c;k++){
				cin>>arr[j][k];
			}
		}
		
		int store=rotten(&arr[0][0],r,c);
		if(!zero(&arr[0][0],r,c))
			cout<<"-1"<<endl;
		else
			cout<<store<<endl;
	}
	
	return 0;
}

Output

Number of days are : 2