Akamai Interview Question for Quality Assurance Engineers


Country: India
Interview Type: In-Person




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1
of 1 vote

You can maintain three variable to store the top 3 elements while sweeping the list.

- wdxcn1123 September 04, 2012 | Flag Reply
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0
of 0 votes

Correct Answer.

- Anonymous September 04, 2012 | Flag
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1
of 1 vote

1. Start one pointer "fp" and iterate through linkedlist until it reaches 3rd node.
2. When "fp" reaches 3rd element, start another pointer "sp" from head node.
3. From now on, advance both pointers"fp" and "sp", node by node until you reach "fp"pointing to NULL.
4. At this time "sp" is pointing to 3rd element from end of the linked list
5. In other words, iterating a pointer to maintain some gap (here 3 in this example)

- NaveenG October 06, 2012 | Flag Reply
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0
of 0 votes

Correct -- Awesome ...Thanks

- Neo February 26, 2013 | Flag
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0
of 0 votes

Thx

- u.ranjith August 05, 2013 | Flag
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0
of 0 votes

/**
       *  find kth element
       *  o(n)
       */
       public void findKthElement(int k)
       {
      	if( head == null ){ //handle k >= length of list
      		return;
      	}
      	Node<AnyType> fast = head;
      	Node<AnyType> slow = head;
      	while(k > 0){
      		fast = fast.next;
      		k--;
      	}
      	while(fast != null){
      		fast = fast.next; 
      		slow = slow.next;	
      	}
      	
      	System.out.println("kth element value is: " + slow.data );
       }

- Algorithmy October 23, 2014 | Flag
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0
of 0 vote

check if the node->next->next->next is null..if so, then you are at the n-3 rd element.

- BJ September 09, 2012 | Flag Reply
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0
of 0 votes

lol no one said that this was the simplest and correct answer lol.

- anon July 28, 2013 | Flag
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0
of 0 vote

I don't think you need 3 temps, 2 are sufficient, one t1 = head->next->next->next, and t2 = head.
Make ++t1, and ++t2, until t1 == NULL.

- nathaniel September 11, 2012 | Flag Reply
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0
of 0 vote

public Link find(Link first)
{
Link current =first;
Link fast=first;
while (fast !=null)
{
fast=fast.next.next.next;
current=current.next;
}
return current;
}

- m September 29, 2012 | Flag Reply
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0
of 0 vote

Hi

- a_NEU November 02, 2014 | Flag Reply
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0
of 0 vote

Node* findNthToLast(Node* head, int n)
{
	Node* temp = head;
	Node* nthToLast = NULL;
	int count = 0;
	while(temp != NULL)
		{
			count++;
			temp = temp->next
			if(count == n)
			{
				nthToLast = head;
			}
			else if (count > n)
			{
				nthToLast = nthToLast ->next;
			}

		}
		if(nthToLast == NULL)
		{
			cout << "Error: List size is smaller than N"<<endl;
		}
		return nthToLast;
}

- XnakelX May 20, 2016 | Flag Reply
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0
of 0 vote

public void getNthMinusyElement(int n){
System.out.println();
ListNode curr = head;
ListNode nthToLast = null;
int count = 0;
while(curr != null)
{
count++;
curr = curr.getNext();
if(count == n)
{
nthToLast = head;
}
else if (count > n)
{
nthToLast = nthToLast.getNext();
System.out.println("data==="+nthToLast.getData());

}

}
System.out.println(nthToLast.getData());

}

- Vipul Agarwal April 08, 2017 | Flag Reply


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