GATE CS 2025 Set 1 Q.65: Hash Tables and Collision Resolution

In a double hashing scheme, $h_1(k) = k \bmod 11$ and $h_2(k) = 1 + (k \bmod 7)$ are the auxiliary hash functions. The size $m$ of the hash table is $11$ . The hash function for the $i$ -th probe in the open address table is $[h_1(k) + i\, h_2(k)] \bmod m$ . The following keys are inserted in the given order: $63, 50, 25, 79, 67, 24$ .

The slot at which key $24$ gets stored is ___________. (Answer in integer)

Your answer

Official answer and solution

Accepted answers: 10 to 11
From: Official GATE 2025 answer key

Solution

Insert the first five keys using $h_1(k) = k \bmod 11$ , then probe $i = 0, 1, 2, \dots$ for key $24$ using $h_1(24) = 2$ and $h_2(24) = 4$ .

Insert the first 5 keys ( $i = 0$ $i = 0$ ):
- $k = 63$ : $h_1(63) = 63 \bmod 11 = 8 \implies$ Slot 8
- $k = 50$ : $h_1(50) = 50 \bmod 11 = 6 \implies$ Slot 6
- $k = 25$ : $h_1(25) = 25 \bmod 11 = 3 \implies$ Slot 3
- $k = 79$ : $h_1(79) = 79 \bmod 11 = 2 \implies$ Slot 2
- $k = 67$ : $h_1(67) = 67 \bmod 11 = 1 \implies$ Slot 1
Insert $k = 24$ :
- $h_1(24) = 24 \bmod 11 = 2$ and $h_2(24) = 1 + (24 \bmod 7) = 1 + 3 = 4$ .
- Probe $i = 0$ : $[2 + 0 \times 4] \bmod 11 = 2$ (occupied by $79$ ).
- Probe $i = 1$ : $[2 + 1 \times 4] \bmod 11 = 6$ (occupied by $50$ ).
- Probe $i = 2$ : $[2 + 2 \times 4] \bmod 11 = 10$ (empty!).

Thus key $24$ is stored at slot 10 (or 11 if 1-indexed; official key accepts 10 to 11).

Final answer

10 (10 to 11 accepted)

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2025 Computer Science and Information Technology - Session 1 (CS1)
Question: 65 of 65
Section: Computer Science and Information Technology

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.

Question

Solution

Source