Optimal game strategy recursion , why is this codde wrong

#1

Scanner sc=new Scanner(System.in);
int N=sc.nextInt();

	ArrayList<Integer> arr= new ArrayList<>();

	for(int i=0 ; i<N ; i++)
	{
		int a=sc.nextInt();
		arr.add(a);
	}

	int ans=ogs(arr,0,arr.size()-1 , arr.size());

	System.out.println(ans);
	
}


static int ogs(ArrayList<Integer> arr , int s , int e , int t)
{

	if(t==1) return 0  /*arr.get(s)*/;
	
	
	int piyush=0,p=0,n=0;
	
	
	
	
			if(arr.get(s)>arr.get(e))
			{		
				if(arr.get(e)>arr.get(e-1))
				{
					if(arr.get(s+1)>arr.get(s))
					{
						piyush+=arr.get(e);
						arr.remove(e);
					}
					else
					{
						piyush+=arr.get(s);
						arr.remove(s);
					}
				}
				else
				{
					if(arr.get(s+1)>arr.get(s))
					{
						piyush+=smaller(  arr.get(s+1)-arr.get(s)  ,   arr.get(e-1)-arr.get(e)  );
						arr.remove(  smaller(  arr.get(s+1)-arr.get(s)  ,   arr.get(e-1)-arr.get(e)  )   ) ;
					}
					else
					{
						piyush+=arr.get(s);
						arr.remove(s);
					}
				}
			}
			else
			{
				if(arr.get(s+1)>arr.get(s))
				{
					if(arr.get(e)>arr.get(e-1))
					{
						piyush+=arr.get(e);
						arr.remove(e);
					}
					else
					{
						piyush+=smaller(  arr.get(s+1)-arr.get(s)  ,   arr.get(e-1)-arr.get(e)  );
						arr.remove(  smaller(  arr.get(s+1)-arr.get(s)  ,   arr.get(e-1)-arr.get(e)  )   ) ;
					}
				}
				else
				{
					if(arr.get(e)>arr.get(e-1))
					{
						piyush+=arr.get(e);
						arr.remove(e);
					}
					else
					{
						piyush+=arr.get(s);
						arr.remove(s);
					}
				}				
			}
	
	
	
		if(t%2==0)
		{
			p=piyush;
			n+=ogs(arr,s,arr.size()-1,arr.size());
			if(t>2)
			p+=n;
		}
		else
		{
			n=piyush;
			p+=ogs(arr,s,arr.size()-1,arr.size());
		}
	
	
		return p;

}


static int smaller(int m,int n)
{
	if(m>n) return n;
	else return m;
#2

@Himanshu-Jhawar-2273952536067590,
Don’t use an arraylist. Use simple recursion.

        Consider both the possibilities. You can pick either the first or the last coin.
        Since the opponent plays optimally , we would get the minimum of the remaining coins for each choice.

        Pick the max of two as your final result
#3

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