Last Updated :
20 Dec, 2018
The keys 123, 164, 232, 38, 195 and 253 are inserted into an empty hash table of size 11 using open addressing with the hash function H(key) = key mod 11. Which slot will remain empty?
(A) 0, 1, 4, 5, 10
(B) 0, 3, 4, 7, 8
(C) 1, 2, 6, 9, 10
(D) 1, 3, 7, 9, 10
Answer: (B)
Explanation:
123 mod 11 = 1
164 mod 11 = 9
232 mod 11 = 1
38 mod 11 = 5
195 mod 11 = 10
253 mod 11 = 5
The hash table will be:
| Slot |
key |
| 0 |
– |
1
|
12
3
|
| 2 |
23
2
|
| 3 |
– |
| 4 |
– |
| 5 |
3
8
|
| 6 |
25
3
|
| 7 |
– |
| 8 |
– |
| 9 |
16
4
|
| 10 |
19
5
|
undefined
So, option (B) is correct.
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