CareerCup

Can you shed some more light on below para?

"After the tree constructed, take root and do all traversal to all possible leaf at level count ( decimal points of start ) and generate a number for such branch."

If I understood correctly, traversing through a big tree will also cost O(h), which is almost equal to O(n). So I guess there might not be much difference asymptotically. Please correct me if I'm wrong.

It would be helpful if you can provide a Pseudocode for your approach.

public class glassdoor2{

public boolean check(int i){
	int prev=i%10;
	int curr;
	i=i/10;
	while (i>0)
	{
		curr=i%10;
		if ((prev-curr==1)||(prev-curr==-1))
			prev=curr;
		else 
			return false;	
			
		i=i/10;	
	}
	
	return true;
}

public void compute(int start,int end)
{
	for(int i=start;i<end+1;i++)
	{
		if(check(i))
			System.out.println(i);
	}
}

public static void main(String[] args){
	glassdoor2 g = new glassdoor2();
	g.compute(1000,400000);
}

asymptotic solution is O(n*d). worst case O(n2).