Answer followed by explanation:
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
24 | 120 | 17 | 78 | 146 | 189 | 114 | 42 | 282 |
Using a similar approach as used in inserting the first 4 elements we can insert the rest of the elements:
120: h(120) = 0. This is filled. Apply double hashing:
h(120) + 1.h2(120) = 0 + 1 = 1 which is empty hence 120 goes to position 1.
78: h(78) = 3 which is empty hence it goes to index 3.
114: h(114) = 9 . Filled:
h(114) + 1. h2(114) = 11 mod 15 = 11. Empty Hence 114 goes to 11.
146: h(146) = 11. Filled
h(146) + h2(146) = 11 + 6 mod 15 = 17 mod 15 = 2. Filled
h(146) + 2. h2(146) = 11 + 12 mod 15 = 23 mod 15 = 8. Empty Hence 146 goes to 8
282: h(282) = 12. Filled
h(282) + 1.h2(282) = 12 + 2 = 14 mod 15 = 14. Empty Hence 282 goes to index 14.
Finally, Hence the table looks like:
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
24 | 120 | 17 | 78 | 146 | 189 | 114 | 42 | 282 |
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