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# GATE | GATE CS 1996 | Question 63

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

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```

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