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Question

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?

A
0
1
2 12
3 13
4
5 5
6
7
8 18
9
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B
0
1
2 12
3 13
4 2
5 3
6 23
7 5
7 18
9 15
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C
0
1
2 12,2
3 13,3,23
4
5 5,15
6
7
8 18
9
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D
0
1
2 2
3 23
4
5 15
6
7
8 18
9
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Solution

The correct option is B
0
1
2 12
3 13
4 2
5 3
6 23
7 5
7 18
9 15

12 mod 10 = 2
18 mod 10 = 8
13 mod 10 = 3
2 mod 10 = 2 collision
(2 + 1) mod 10 = 3 again collision (using linear probing)
(3 + 1) mod 10 = 4
3 mod 10 = 3 collision
(3 + 1) mod 10 = 4 collision (using linear probing)
(4 + 1) mod 10 = 5
23 mod 10 = 3 collision
(3 + 1) mod 10 = 4 collision
(4 + 1) mod 10 = 5 again collision
(5 + 1) mod 10 = 6
5 mod 10 = 5 collision
(5 + 1) mod 10 = 6 again collision
(6 + 1) mod 10 = 7
15 mod 10 = 5 collision
(5 + 1) mod 10 = 6 collision
(6 + 1) mod 10 = 7 collision
(7 + 1) mod 10 = 8 collision
(8 + 1) mod 10 = 9

So, resulting hash table
0
1
2 12
3 13
4 2
5 3
6 23
7 5
8 18
9 15

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