Insert the characters of the string K R P C S N Y T J M into a hash table of size 10.
Use the hash function
h(x) = ( ord(x) – ord("a") + 1 ) mod10
If linear probing is used to resolve collisions, then the following insertion causes collision
(A) Y
(B) C
(C) M
(D) P
Answer: (C)
Explanation:
(a) The hash table with size 10 will have index from 0 to 9.
hash function = h(x) = ((ord(x) – ord(A) + 1)) mod 10
So for string K R P C S N Y T J M:
K will be inserted at index : (11-1+1) mod 10 = 1
R at index: (18-1+1) mod 10 = 8
P at index: (16-1+1) mod 10 = 6
C at index: (3-1+1) mod 10 = 3
S at index: (19-1+1) mod 10 = 9
N at index: (14-1+1) mod 10 = 4
Y at index (25-1+1) mod 10 = 5
T at index (20-1+1) mod 10 = 0
J at index (10-1+1) mod 10 = 0 // first collision occurs.
M at index (13-1+1) mod 10 = 3 //second collision occurs.
Only J and M are causing the collision.
(b) Final Hash table will be:
0 T 1 K 2 J 3 C 4 N 5 Y 6 P 7 M 8 R 9 S
