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1. A hash table of length 10 uses open addressing with hash function 0 h(k)-k mod 10, and linear probing. After inserting 6 values into

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1. A hash table of length 10 uses open addressing with hash function 0 h(k)-k mod 10, and linear probing. After inserting 6 values into an empty hash table, the table is as shown below. Which one of the following choices gives a possible order in which the key values3 23 42 could have been inserted in the table? 4 34 5 52 6 46 A. 33, 42, 34, 52, 23, 46 B. 34, 23, 42, 52, 46, 33 C. 52, 34, 42, 23, 46, 33 D. 42, 34, 33,23, 52,46 2. The keys 12, 18, 15, 2, 3, 23, and 5 are inserted into an initially empty hash table of length 10 using open addressing with hash function h(k) = k mod 10 and linear probing. What is the resultant hash table? (Fill in the table on the right.) 3. Solve the question 2 with a chained hash table (external hashing) over the open addressing scheme. Fill in the table below

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