Common Cause

TIP102 Unit 1 Session 2 Standard (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Easy
• ⏰ Time to complete: 10 mins
• 🛠️ Topics: Arrays, Sets, Iteration

U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?

• The function common_elements() should take two lists lst1 and lst2 and return a list of elements that are present in both lists.

HAPPY CASE
Input: lst1 = [""super strength"", ""super speed"", ""x-ray vision""], lst2 = [""super speed"", ""time travel"", ""dimensional travel""]
Expected Output: [""super speed""]

Input: lst1 = [""apple"", ""banana"", ""cherry""], lst2 = [""banana"", ""cherry"", ""date""]
Expected Output: [""banana"", ""cherry""]

EDGE CASE
Input: lst1 = [""super strength"", ""super speed"", ""x-ray vision""], lst2 = [""martial arts"", ""stealth"", ""master detective""]
Expected Output: []

Input: lst1 = [], lst2 = [""time travel""]
Expected Output: []

P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Convert one list to a set for O(1) average time complexity checks and iterate through the other list to find common elements.

1. Convert `lst2` to a set called `set2` for faster lookups.
2. Initialize an empty list `result` to store common elements.
3. Iterate through each element in `lst1`:
   a. If the element is in `set2`, append it to `result`.
4. Return `result`

⚠️ Common Mistakes

• Not accounting for duplicates in the input lists.
• Not handling cases where one or both lists are empty.

I-mplement

Implement the code to solve the algorithm.

def common_elements(lst1, lst2):
    set2 = set(lst2)
    result = []
    for item in lst1:
        if item in set2:
            result.append(item)
    return result