Set .discard(), .remove() & .pop()- HackerRank Solution

 Set .discard(), .remove() & .pop()- HackerRank Solution

.remove(x)

This operation removes element  from the set.
If element  does not exist, it raises a KeyError.
The .remove(x) operation returns None.

Example

>>> s = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
>>> s.remove(5)
>>> print s
set([1, 2, 3, 4, 6, 7, 8, 9])
>>> print s.remove(4)
None
>>> print s
set([1, 2, 3, 6, 7, 8, 9])
>>> s.remove(0)
KeyError: 0

---

.discard(x)

This operation also removes element  from the set.
If element  does not exist, it does not raise a KeyError.
The .discard(x) operation returns None.

Example

>>> s = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
>>> s.discard(5)
>>> print s
set([1, 2, 3, 4, 6, 7, 8, 9])
>>> print s.discard(4)
None
>>> print s
set([1, 2, 3, 6, 7, 8, 9])
>>> s.discard(0)
>>> print s
set([1, 2, 3, 6, 7, 8, 9])

---

.pop()

This operation removes and return an arbitrary element from the set.
If there are no elements to remove, it raises a KeyError.

Example

>>> s = set([1])
>>> print s.pop()
1
>>> print s
set([])
>>> print s.pop()
KeyError: pop from an empty set

---

Task

You have a non-empty set , and you have to execute  commands given in  lines.

The commands will be pop, remove and discard.

Input Format

The first line contains integer , the number of elements in the set .
The second line contains  space separated elements of set . All of the elements are non-negative integers, less than or equal to 9.
The third line contains integer , the number of commands.
The next  lines contains either pop, remove and/or discard commands followed by their associated value.

Constraints

Output Format

Print the sum of the elements of set  on a single line.

Sample Input

9
1 2 3 4 5 6 7 8 9
10
pop
remove 9
discard 9
discard 8
remove 7
pop 
discard 6
remove 5
pop 
discard 5

Sample Output

4

Explanation

After completing these  operations on the set, we get set. Hence, the sum is .

Note: Convert the elements of set s to integers while you are assigning them. To ensure the proper input of the set, we have added the first two lines of code to the editor.

Set .discard(), .remove() & .pop() Solution

#For beginner
n = int(input())
s = set(map(int, input().split()))
for i in range(int(input())):
    choice=input().split()
    if choice[0]=="pop" :
        s.pop()
    elif choice[0]=="remove" :
        s.remove(int(choice[1]))
    elif choice[0]=="discard" :
        s.discard(int(choice[1]))
print(sum(s))
#Modified version
n = int(input())
s = set(map(int, input().split())) 
for i in range(int(input())):
    eval('s.{0}({1})'.format(*input().split()+['']))
print(sum(s))

Explanation-  We have earlier used sets in many HackerRank problems using the earlier learned concepts we will declare a set s the choice is a list used to take the input. If the index 0 of the list is pop we will pop an element from s or else for remove and discard we will get the location present at index 1 of the list.

In the other code, we have used eval function which evaluates the expression if it is a legal python statement.

To read about sets follow this link-  https://www.geeksforgeeks.org/sets-in-python/