Question

The following values are to be stored in a Hash-Table: 

25, 42, 96, 101, 102, 162, 197, 201

Use the Division method of Hashing with a table size of 11. Use the sequential method for resolving collision. 

Answer 1

The given values are as follows: - 

25 42 96 101 102 162 197 201

Table size is = 11 

Division method of Hashing: -

H (k) = {Hash address range from 0 to m-1. Key (mod) table-size.} 

H (25)= 3 

H (42) = 9 

H (96) = 8 

H (101) = 2 

H (102) = 3 

H (162) = 8 

H (197) =10 

H (201) = 3

The Hash table is as follows

Hash [0] = [197] 

Hash [1] = [NULL] 

Hash [2] = [101]  

Hash [3] = [25] ← Collision Occurred

Hash [4] = [102] ← Inserted in next available slot

Hash [5] = [201] 

Hash [6] = [NULL] 

Hash [7] = [NULL] 

Hash [8] = [96] 

Hash [9] = [42] 

Hash [10] = [162]