Comments for Arden Dertat

Comment on Programming Interview Questions 19: Find Next Palindrome Number by Arda Antikacioglu

Arda Antikacioglu — Mon, 05 Dec 2011 21:30:59 +0000

Actually, this takes O(log(n)) time since mirroring takes as long as the length of n, which is still linear in the length of the input and can be proven to be optimal.

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Comment on Programming Interview Questions 7: Binary Search Tree Check by Arden

Arden — Thu, 01 Dec 2011 19:07:59 +0000

You’re right, changed it. Thanks for the comment..

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Comment on Programming Interview Questions 7: Binary Search Tree Check by Anonymous

Anonymous — Thu, 01 Dec 2011 04:54:58 +0000

There’s no reason to build a list of all node values in isBST2. You only need the value of the last node visited so far. Rename nodes to maxValBox and update it like this:

maxValBox[0] = tree.val

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Comment on Programming Interview Questions 8: Transform Word by Ilke

Ilke — Wed, 30 Nov 2011 06:07:26 +0000

I didn’t suggest loading the dictionary into a trie and then using it as it were graph :)
Yes, if you use a Trie, every time you pop an element from the queue, you have to find all of its one-step-away neighbors on the fly, however this lookup is very cheap in a trie. Furthermore, you don’t have to do full lookups for every possible neighboring word for additions / changes (for removal you have to, but that’s only length – 1 lookups), the trie will guide you instead with non-null pointers.
English language has more than 1 million words and it’s expanding by ~10k words each year. Plus, source and target word inputs are bounded by the dictionary. Yes, this solution’s lookups are faster on paper, but because of these two facts, I would go with a trie based solution rather than creating that huge hashtable, especially because it would be more efficient in practice.

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Comment on Programming Interview Questions 5: Linked List Remove Nodes by Arden

Arden — Tue, 29 Nov 2011 17:05:08 +0000

Great solution, thanks!

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Comment on Programming Interview Questions 8: Transform Word by Arden

Arden — Tue, 29 Nov 2011 16:32:20 +0000

Trie won’t work because we can change or remove characters. For example, in a trie there won’t be a direct link from cat to at, but we need that link here. And there won’t be a link between cat and bat as well. Also in a trie intermediate nodes are not guaranteed to be actual words. Since trie is a tree it can’t contain backward links, so it would work only if we were allowed to add characters, without removing or changing them.

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Comment on Programming Interview Questions 8: Transform Word by Ilke

Ilke — Tue, 29 Nov 2011 08:24:17 +0000

Hey,
You may want to load the dictionary into a Trie here, rather than a graph as you described.

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Comment on Programming Interview Questions 5: Linked List Remove Nodes by Anonymous

Anonymous — Sun, 27 Nov 2011 09:25:08 +0000

The following pattern removes the need for special cases.

Node* RemoveNodesWithValue(Node* head, int k) { if (!head) { return head; }


	Node dummy;

	dummy.next = head;

	Node* prev = &dummy;

	Node* current = head;
	while (current) {

		if (current->data == k) {

			prev->next = current->next;

			delete current;

			current = prev->next;

		}

		else {

			prev = current;

			current = current->next;

		}

	}

return dummy.next; }

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Comment on Programming Interview Questions 6: Combine Two Strings by Arden

Arden — Sun, 27 Nov 2011 04:16:39 +0000

That’s a great idea, we can definitely do that. We will save so much space by caching the indexes instead of the whole substring itself. And we can easily implement it by slightly modifying the above algorithm. Thanks for the idea..

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Comment on Programming Interview Questions 6: Combine Two Strings by hero

hero — Sat, 26 Nov 2011 23:22:17 +0000

Hi Arden,

I wonder if it would be possible to keep only the index number of failing string compared to original one in the cache instead of whole substring. For your example, it would cache the string tuple as {bc,f}, for some time rigth? Instead, would it be possible to cache {1,2} meaning that 1 is the substring of abc from the index 1 and so on.

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