Sets- discard(), remove() and pop() operations | HackerRank Solutions

 In this problem we  given size of set and space separated elements and  size of operations that we need to perform over set elements , we have to handle here multiple operations and perform and at end return sum of all elements  that  left out inside the set  and if no  elements left then return just  0 

below solutions is kind naive approach not fully  optimized or better but you can do that steps my step

you have to  map  each operation and handle its

 

Problem Breakdown

The task involves:

1. Reading the size of the set and the elements within it.

2. Handling a list of operations that can modify the set.

3. Performing the operations one by one.

4. At the end, returning the sum of the remaining elements in the set, or 0 if the set is empty.

The operations you need to handle include:

• pop: Removes and returns an arbitrary element from the set.

• remove: Removes a specific element from the set. If the element is not found, an error is raised.

• discard: Removes a specific element from the set if it exists. If the element is not found, nothing happens (no error is raised).

In this naive solution, we iterate over the operations and modify the set according to the instructions. The operations are read as strings, split into a list, and then executed based on the operation type.

Problem:https://www.hackerrank.com/challenges/py-set-discard-remove-pop/problem?isFullScreen=true

Code:

#Naive Solutions:

n = int(input())

s = set(map(int, input().split()))

#naive approcah

m = int(input())

for _ in range(m):

    x = input().split()

    if x[0] == 'pop':

        s.pop()

    elif x[0] == 'pop' and x[1]:

        s.pop(int(x[1]))

    elif x[0] == 'remove':

        s.remove(int(x[1]))

    else:

        s.discard(int(x[1]))

if len(s)==0:

    print(0)

else:

    print(sum(s))

Time Complexity : Big(O)n

O(n)

The code snippet has a loop that runs 'm' times, where 'm' is the input value. Inside the loop, there are operations like pop, remove, and discard on the set 's'. The worst-case time complexity of these operations is O(1). Therefore, the overall time complexity of the code is O(m), where 'm' is the input value.

1. Input Parsing:

   • We first read the size of the set n (although it's not really necessary for the solution).

   • We then create a set s by mapping the input elements to integers.

   • Next, we read the number of operations m that we need to perform on the set.

2. Operation Handling:

   • The loop iterates m times, processing each operation.

   • For each operation:

     • pop: Removes a random element from the set.

     • remove: Removes a specified element from the set. If the element does not exist, it raises an error (but we assume the input ensures the element is present).

     • discard: Removes a specified element but does nothing if the element is not found.

3. Final Output:

   • After all operations are performed, we check if the set is empty.

   • If the set is empty, we print 0; otherwise, we print the sum of the remaining elements in the set.

Key Concepts:

• Set Operations: Understanding the difference between pop, remove, and discard is crucial when solving this problem. pop removes a random element, remove throws an error if the element is not found, and discard silently handles missing elements.

• Efficiency Considerations: While this solution is straightforward, there are optimizations that could be made for more complex scenarios (e.g., handling invalid operations gracefully or reducing the complexity of repeated set modifications).

This problem is a great exercise in handling basic set operations in Python. The operations themselves—pop, remove, and discard—are common in Python and understanding their differences is key to solving problems that involve sets.