Question

Could someone give me the right answer to this question and explain it please. Thank you.

If a hash table with 7 entries (hash code 0 to 6) uses linear probing, which entry is the last to be filled when values are inserted according to this hash sequence:

0 1 4 3 1 2 0

Select one:

0

2

6

5

Answer 1

0 is the last element that is entered into the hash table.

Answer to this question can be understood by understanding how duplicate elements are treated in the linear probing or any hash algorithm. Each element is passed through a hash function and based on its output its position is decided in the hash table. So basically a duplicate element is bound to have a same table location. What happens here is, the previous element is replaced by the new one. It doesn't matter much as the elements are same. And whenever a search operation is performed it will return the same address (input can't differentiate whether the first or the next element is input, its same for the compiler as values are the same).

Eg: in our program lets assume the hash function gives the element itself i.e. h(x) = x

0: | 0 | | | | | | |

1: | 0 | 1 | | | | | |

4:   | 0 | 1 | | | 4 | | |

3:   | 0 | 1 | | 3 | 4 | | |

1:   | 0 | 1 | | 3 | 4 | | | (The old one is replaced by the new one)

2:   | 0 | 1 | 2 | 3 | 4 | | |

0:   | 0 | 1 | 2 | 3 | 4 | | |

What should've happened here when the second 1 was inserted into the table was, it should've gone to position 2 as it was the next empty slot. But in that case there during search operation it would've returned two values which would be wrong.