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The keys $12, 18, 13, 2, 3, 23, 5$ and $15$ are inserted into an initially empty hash table of length $10$ using open addressing with hash function $h(k) = k \mod 10$ and linear probing. What is the resultant hash table?

1. $\begin{array}{|l|l|} \hline 0 & \\\hline 1 & \\\hline 2 & 2 \\\hline 3 & 23 \\\hline 4 & \\\hline 5 & 15 \\\hline 6 & \\\hline 7 & \\\hline 8 & 18 \\\hline 9 & \\\hline \end{array}$
2. $\begin{array}{|l|l|} \hline 0 & \\\hline 1 & \\\hline 2 & 12 \\\hline 3 & 13 \\\hline 4 & \\\hline 5 & 5 \\\hline 6 & \\\hline 7 & \\\hline 8 & 18 \\\hline 9 & \\\hline \end{array}$
3. $\begin{array}{|l|l|} \hline 0 & \\\hline 1 & \\\hline 2 & 12 \\\hline 3 & 13 \\\hline 4 & 2 \\\hline 5 & 3 \\\hline 6 & 23 \\\hline 7 & 5 \\\hline 8 & 18 \\\hline 9 & 15 \\\hline \end{array}$
4. $\begin{array}{|l|l|} \hline 0 & \\\hline 1 & \\\hline 2 & 2,12 \\\hline 3 & 13,3,23 \\\hline 4 & \\\hline 5 & 5,15 \\\hline 6 & \\\hline 7 & \\\hline 8 & 18 \\\hline 9 & \\\hline \end{array}$
in DS
edited | 1.4k views

(C) is the correct option ..directly from the definition of linear probing. In linear probing, when a hashed location is already filled, locations are linearly probed until a free one is found.

http://courses.cs.washington.edu/courses/cse326/00wi/handouts/lecture16/sld015.htm

by Boss (14.3k points)
edited by
+8
Note:

Separate Chaining  = Open Hashing/Closed Addressing (i.e. each key has a linked list, hence no probing)

Closed Hashing/Open Addressing (single array shared, hence probing required)
Here A & B options are incorrect, this is obvious. We are inserting 8 keys but only 4 are present. Hash is data structure for storing data, we don't loose data in Hash.

D is incorrect because it looks like chaining .

Using Linear Probing we get hash table of C. In linear probing, when a hashed location is already filled, locations are linearly probed until a free one is found.
by Boss (41.2k points)

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