Consider a hash table with 9 slots. The hash function is $h(k) = k \bmod 9$. The collisions are resolved by chaining. The following 9 keys are inserted in the order: 5, 28, 19, 15, 20, 33, 12, 17, 10. The maximum, minimum, and average chain lengths in the hash table, respectively, are

We are given a hash table with $m = 9$ slots (indices $0$ to $8$) and the hash function $h(k) = k \bmod 9$.

Let us compute the slot index for each of the 9 inserted keys:

• $h(5) = 5 \bmod 9 = 5$

• $h(28) = 28 \bmod 9 = 1$

• $h(19) = 19 \bmod 9 = 1$

• $h(15) = 15 \bmod 9 = 6$

• $h(20) = 20 \bmod 9 = 2$

• $h(33) = 33 \bmod 9 = 6$

• $h(12) = 12 \bmod 9 = 3$

• $h(17) = 17 \bmod 9 = 8$

• $h(10) = 10 \bmod 9 = 1$

Distribution of keys in the slots:

• Slot 0: None $\implies \text{length} = 0$

• Slot 1: $28, 19, 10 \implies \text{length} = 3$

• Slot 2: $20 \implies \text{length} = 1$

• Slot 3: $12 \implies \text{length} = 1$

• Slot 4: None $\implies \text{length} = 0$

• Slot 5: $5 \implies \text{length} = 1$

• Slot 6: $15, 33 \implies \text{length} = 2$

• Slot 7: None $\implies \text{length} = 0$

• Slot 8: $17 \implies \text{length} = 1$

From this distribution:

• Maximum chain length: $3$ (in slot 1)

• Minimum chain length: $0$ (in slots 0, 4, 7)

• Average chain length: $\frac{\text{Total number of keys}}{\text{Number of slots}} = \frac{9}{9} = 1$

Therefore, the values are 3, 0, and 1, which corresponds to option (A).