Amazon Interview Question for Software Developers


Country: United States
Interview Type: Phone Interview




Comment hidden because of low score. Click to expand.
1
of 1 vote

Here is how we will do this in two moves.

public static Node DeepCopy(Node root)
{
	if(root == null)
	{
		return null;
	}
	Node  currentNode = root;
	// this loop will create all the new nodes in the same input linkedlist.
	while(currentNode != null)
	{
		Node temp = new Node(currentNode.data);
		temp.next = currentNode.next;
		currentNode.next = temp;
		currentNode = temp.next;
	}
	currentNode = root;
	Node newList = currentNode.next;
	Node newCurrentNode = currentNode.next;
	while(currentNode != null)
	{
		newCurrentNode = currentNode.next;
		if(currentNode.random != null)
		{
			newCurrentNode.random = currentNode.random.next;
		}
		currentNode.next = newCurrentNode.next;
		currentNode = newCurrentNode.next;
		newCurrentNode.next = currentNode.next;
	}
	return newList;
}

Complexity is O(n) + O(1).

- sonesh April 20, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I came up with below answer after modifying the Node class. The complexity is O(n) even though I might have to go through the entire list thrice. Correct me on complexity if I am wrong.

public class Node
{
	public int Value {get;set;}
	public Node Next{get;set;}
	public Node Random{get;set;}
	public int NodeCountFromHead {get;set;}
}

static Node DeepCopy(Node sourceHead)
{
	if(sourceHead == null)
		return null;
	
	int index = 1;
	sourceHead.NodeCountFromHead = index;
	
	Node sourceCurrent = sourceHead;
	Node destHead = new Node {Value = sourceHead.Value, Next = null};
	Node destCurrent = destHead;
	
	while(sourceCurrent.Next != null)
	{
		sourceCurrent.Next.NodeCountFromHead = ++index;
		destCurrent.Next = new Node{Value = sourceCurrent.Next.Value, Next = null, Random = null, NodeCountFromHead = index};
		sourceCurrent = sourceCurrent.Next;
		destCurrent = destCurrent.Next;
	}
	
	// second loop to set the Random node values
	sourceCurrent = sourceHead;
	destCurrent = destHead;
	while(sourceCurrent != null)
	{
		destCurrent.Random = GetNthNode(destHead, sourceCurrent.Random.NodeCountFromHead);
		sourceCurrent = sourceCurrent.Next;
		destCurrent = destCurrent.Next;
	}
	
	return destHead;
}

static Node GetNthNode(Node destHead, int nodeCountFromHead)
{
	Node current = destHead;
	
	int index = 1;
	while (index != nodeCountFromHead)
	{
		current = current.Next;
	} 
	
	return current;
}

- Abcd April 20, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

That's great. Thank you!

Only one question: O(n) makes sense. Where did you get O(1) from?

- Abcd April 21, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I generally write time and space complexity together. So O(n) is time and O(1) extra space apart from what we have to use.

- sonesh April 21, 2017 | Flag Reply


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