Comments on: Programming Interview Questions 15: First Non Repeated Character in String

By: Kowshik

Kowshik — Sun, 18 Dec 2011 09:36:33 +0000

An optimization which a friend of mine pointed out to me: Instead of an array of pairs, we can use just use an array of integers, where -1 (default value) represents non-occurence of the character and -2 represents duplication.

]]>

By: Arden

Arden — Sat, 17 Dec 2011 17:54:03 +0000

You’re totally right. I have a post ready to publish discussing exactly this solution. It’ll be number 25: remove duplicate characters in string. Thanks for the comment..

]]>

By: Kowshik

Kowshik — Sat, 17 Dec 2011 11:02:27 +0000

Heres a twist to the same problem that I encountered in one of my recent interviews. What if the given string is extremely huge and the character set is lowercase letters (‘a’ to ‘z’)?

Answer: Firstly, we can replace the hash table with an array of pairs of size 26 (each representing a lowercase letter):
– The first element in each pair stores the location where the character occured for the first time in the string (default value is -1 indicating non-occurence).
– The second element in each pair is a boolean that indicates if the character’s occurence is unique. (default value is false and it also indicates repetition. true indicates uniqueness.)

With this data structure ready, it is enough if we parse through the original string only once to populate the array of pairs. Then we do a scan on the array to figure out the first unique character.

]]>

By: Arden

Arden — Mon, 14 Nov 2011 20:48:30 +0000

Good alternative solution. But if we replace the binary search tree with hashtable, updating the counts will be slightly faster I guess?

]]>

By: alp

alp — Mon, 14 Nov 2011 20:21:13 +0000

I think we can traverse the string only once by keeping two trees, one is just a binary search tree to keep nodes of characters and their counts, one is a minheap to keep those nodes sorted and while traversing we’ll maintain the heap and then after the traversal is finished, answer is at the root of the heap. That comes with a memory paste and we can maintain the heap in O(N) time, I think. That’s not a better solution, just wanted to leave it here.

]]>