MADE EASY TEST SERIES

Question

Consider a hash table of size 11. Let h(k) = k mode 11 be the hash function used. A sequence of records with keys {33, 68, 62, 47, 48, 60, 104, 23, 19, 120, 87} is inserted into an initially empty hash table, the bins of which are indexed from zero to ten.

If collisions are resolved by linear probing then maximum number of collisions in worst case will be?

Comments

h(k) = k mode 11
h(33) = 33 mode 11 = 0
h(68) = 68 mode 11 = 2
h(62) = 62 mode 11 = 7
h(47) = 47 mode 11 = 3
h(48) = 48 mode 11 = 4
h(60) = 60 mode 11 = 5
h(104) = 104 mode 11 = 5
acrdng to linear probing it will move to next position 6 which is empty
so max collsion is 1
h(23) = 23 mode 11 = 1
h(19) = 19 mode 11 = 8
h(120) = 120 mode 11 = 10
h(87) = 87 mode 11 = 10
it will cause collsion and move to next position upto 9th position causing 10 collsions
so max collision = 11 

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But I am getting 10 collisions in worst case because 87 mod 11 = 10 and therefore it will hash to index position 10.

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I corrected it

Answer 1

Max. Collision possible = 1

Comments

How? I know this question is very simple but I am not able to understand what the question is actually asking us to find. Please guide.