GATE IT 2008 | Question: 48

Question

Consider a hash table of size $11$ that uses open addressing with linear probing. Let $h(k) = k \mod 11$ be the hash function used. A sequence of records with keys

$43 \ 36 \ 92 \ 87 \ 11 \ 4 \ 71 \ 13 \ 14$

is inserted into an initially empty hash table, the bins of which are indexed from zero to ten. What is the index of the bin into which the last record is inserted?

• A. $3$
• B. $4$
• C. $6$
• D. $7$

Answer 1

$$\begin{array}{|c|c|} \hline \textbf{Index} &  \textbf{key}\\\hline 0 & 87 \\\hline 1 & 11 \\\hline 2 & 13 \\\hline 3 & 36 \\\hline 4 & 92 \\\hline 5 & 4 \\\hline 6 & 71 \\\hline \textbf{7} & \textbf{14} \\\hline 8 \\\hline 9 \\\hline 10 & 43 \\\hline  \end{array}$$

(D) is answer

Answer 2

Answer: D

The hash table looks like this at the end:

87,11,13,36,92,4,71,14,-,-,43

Answer 3

Answer D

Index	Key
0	87
1	11
2	13
3	36
4	92
5	4
6	71
7	14
8
9
10	43

Comments

whats the difference when its said open addressing or closed addressing
?

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Separate Chaining  = Open Hashing
Open Addressing = Closed Hashing

Answer 4

https://youtu.be/q42Dgt0r0c0