I want to keep track of the number of collisions I have. So far, I have observed 2 collisions. No items have been removed from the hash table. I add a single item Charlie to the hash table. The hash position for Charlie is 4. What will my updated values be?Note: We are using Linear Probing to handle collision resolution.

Question

BlackTom AI · Accepted Answer

First, restate the scenario: The hash table uses linear probing for collision resolution, and you have already observed 2 collisions before inserting Charlie. The hash position for Charlie is 4. The current table (based on the diagram) shows that positions 4 and 5 are occupied (Anne at 4, Frank at 5), while 6 is empty.

Option a: '3 collisions' would imply only one additional collision occurred during the insertion. However, with linear probing, Charlie attempts slot 4 first and finds it occupied (collision #1), then moves to slot 5, which is also occupied (collision #2). Only after probing 6 would Charlie be inserted if 6 is empty. Since there are two pre-filled consecutive slots starting at 4, Charlie experiences two collisions during insertion, not one.

Option b: '2 collisions' suggests that inserting Charlie adds two collisions in total, or that there were only two collisions in the entire process after insertion. Given there were already 2 prior collisions, and Charlie encounters two more occupied slots before finding an empty slot, the total becomes 4 collisions, not just 2. Therefore this is incorrect.

Option c: 'It’s impossible to tell' is inaccurate because the diagram and the description provide enough information: we know which slots are occupied and Charlie’s hash index, so we can trace the probing sequence and count collisions.

Option d: '4 collisions' aligns with the step-by-step accounting: initial 2 collisions plus 2 additional collisions encountered during Charlie’s insertion (slots 4 and 5 are occupied), totaling 4. Charlie would be placed in slot 6 once an empty slot is found, but the question focuses on the updated collision count, which is 4.

In summary, linear probing checks index 4 (collision #1), then index 5 (collision #2), then inserts at index 6 (empty). The cumulative collision count becomes 4, matching option d.

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类似问题

Open Addressing One of the advantages of chained hash table over the open addressing scheme is ______

Open Addressing One of the advantages of chained hash table over the open addressing scheme is ______

What is the worst-case complexity of get in separate chaining? (Make no assumption about the load factor or how good the hash function is.)

What is the worst-case complexity of insert in separate chaining? (Make no assumption about the load factor or how good the hash function is.)

In a separate chaining hashtable, what is the impact of increasing the load factor above 1?

Consider the hash table of size 11 that uses linear probing shown below, where the hash function is h(k) = (k mod 11). Thus h(14)=3, h(25)=3. What indices are examined if we now perform get(4)?

Consider an implementation of a hash table with linear probing where the hash function runs in constant time and the cost of comparison is also constant. What is the best-case time complexity of the __getitem__ method? Consider n is the number of items in the hash table.

Which is TRUE?

更多留学生实用工具

希望你的学习变得更简单

Consider an implementation of a hash table with linear probing where the hash function runs in constant time and the cost of comparison is also constant. What is the best-case time complexity of the getitem method? Consider n is the number of items in the hash table.