Question

10. (5 points) Consider data with integer keys 28, 21, 11, 47, 36, 19, 32 in that order inserted into a hash table of size 7 and hashing function is h(key) = k % 7. Show a chaining hash table after doing the insertions:

Answer 1

. Answer ; Insent 28 hlkey)= k % 7 = 1877 = 0 So 28 will stone in o index. 28 Insent qui h(21) = 21%7= 0 . . So ai will store in othindex. Because it chining hashi table. : is T2345 Tail Insert ! h (u) = 11%7= 4. → So ll will stone in the index. 01 13 14 15 16

Inseant 47:- ..h(4-7) – 47%9 = 5 So 47 will stone in oth index oli 12 | 3 | 4 11 g | 47 | In seat 36: 6(36) = 367. 7 = 1 →S0 36 will stone in ist index. 28 36 Inseant 19:- : (19) = 19 %7=5. → So 19 will store in 5th index. In 5th index already contain value. So it will using chaining hash table, so it will chain to 19. 0 28 12 13 14 15 47 36

Inseat 39: h(32) = 32%7=4 → So 4th index already have a element. So it will chain to 3d. Because it is chaining hash table Lo l a | 3 | 4 1.5 T!! 147

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