Answer:
Key = k mod 10
for 13
=13 mod 10 =3
for 44 =44 mod 10 =4
for 103 = 103 mod 10 =3 which collides
hence, we use quadratic probing (u + i ^2) mod 10 for i = 0 to m-1
=(103 + 0)mod 10 =3
=(103 +1) mod 10 =4
=(103 +4) mod 10 =7
Now for 113
= (113+ 0) mod 10 =3
=(113 + 1) mod 10 = 4
=(113 +4) mod 10 =7
=(113 + 9) mod 10 = 1
for 2
= 2 mod 10 =2
so position filled are =[ -, 113, 2, 13, 44,-. -, 103,-.-]
Now we need to search for 33 and 22
= list.contains(33), and this will probe in one go, and will return false
= list.contains(22) will also probe in one go, and will return false.
In fact hashset has response time O(1), and hence it searches the two numbers in one go, and returns true or false.
Hash set has unique values always, and hence there will be no collision.
Explanation:
Please check the answer section.
