1020.Number_of_Enclaves.cpp

/*
    written by Pankaj Kumar.
    country:-INDIA
*/
typedef long long ll ;
typedef vector<int> vl;
typedef vector<vector<int>> vvl;
#define speed cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(0);
/*  Abbrevations  */
#define ff first
#define ss second
#define mp make_pair
#define pb push_back
// Some print
#define no cout<<"NO"<<endl;
#define yes cout<<"YES"<<endl;
// sort
#define all(V) (V).begin(),(V).end()
#define srt(V) sort(all(V))
#define srtGreat(V) sort(all(V),greater<ll>())
// some extra
#define printv(v) for(ll i=0;i<ll(v.size());i++){cout<<v[i]<<" ";} cout<<endl;
#define precision(x) cout<<fixed<<setprecision(x);
#define sz(V) ll(V.size())


/* ascii value
A=65,Z=90,a=97,z=122
*/

/* Some syntax 
//Syntax to create a min heap for priority queue
//priority_queue <int, vector<int>, greater<int>>pq;
*/
/*  --------------------MAIN PROGRAM----------------------------*/
// to run ctrl+b
const ll INF=1e18;
const ll mod1=1e9+7;
const ll mod2=998244353;
// Techniques :
// divide into cases, brute force, pattern finding
// sort, greedy, binary search, two pointer
// transform into graph

// Experience :
// Cp is nothing but only observation and mathematics.


//Add main code here

class Solution {
public:
    void dfs(vector<vector<int>>& arr,int i,int j){
        if(i>=arr.size() or i<0 or j<0 or j>=arr[0].size() or arr[i][j]==0 or arr[i][j]==2)
            return;
        
        arr[i][j]=2;
        
        dfs(arr,i,j+1);
        dfs(arr,i,j-1);
        dfs(arr,i+1,j);
        dfs(arr,i-1,j);
    }
    int numEnclaves(vector<vector<int>>& arr) {
        int ans=0;
        for(int i=0;i<arr.size();i++){
            if(i==0 or i==arr.size()-1){
                for(int j=0;j<arr[i].size();j++){
                    if(arr[i][j]==1){
                        dfs(arr,i,j);
                    }
                }
            }
            
            if(arr[i][0]==1){
                dfs(arr,i,0);
            }
            
            if(arr[i][arr[i].size()-1]==1){
                dfs(arr,i,arr[i].size()-1);
            }
        }
        for(int i=0;i<arr.size();i++){
            for(int j=0;j<arr[i].size();j++){
                if(arr[i][j]==1){
                    ans++;
                }
            }
        }
        
        return ans;
    }
};


/* -----------------END OF PROGRAM --------------------*/
/*
* stuff you should look before submission
* constraint and time limit
* int overflow
* special test case (n=0||n=1||n=2)
* don't get stuck on one approach if you get wrong answer
*/