Consider the following hash function h(x) = x mod 11 with linear probing as the collision handling strategy, and the current table as 0 1 2 3 4 5 6 7 8 9 10 1 24 14 12 16 28 7 31 Do the following: (a) Insert x = 9 into the hash table. Show the steps. (b) Insert x = 42 into the hash table. Show the steps. (c) Find x = 9 in the hash table. Show the steps. (d) Find x = 42 in the hash table. Show the steps. (e) Delete x = 9 from the hash table using lazy deletion. Show the steps. (f) Find x = 42 in the hash table. Show the steps. (g) Explain what would happen if x = 9 was deleted by leaving an empty cell, instead of lazy deletion (leaving a mark). Can we find x = 42 if this was the case