Opportunity to Join A2SV Generation 5: August Conversion Applications Now Open

A2SV is offering a limited number of seats to students who possess potential and dedication. This is your chance to join us and be part of our Generation 5! While our main intake period is in November, marking the start of the academic year, we are pleased to announce that we have a few additional seats available for talented students like yourself.

📅 Application registration closes on Sunday, August 18, 2024, at 11:59 PM EAT.

🎓 Eligibility:
Open to all current students from Addis Ababa University (AAU), Addis Ababa Science and Technology University (AASTU), and Adama Science and Technology University (ASTU), regardless of batch or department.

✨ Requirements:
❄️ Familiarity with Python
❄️ Experience with interview preparation platforms like LeetCode and Codeforces
❄️ Completion of a minimum of 300 problems across LeetCode and Codeforces
❄️ At least 80 active days on LeetCode and Codeforces
❄️ Covered the following topics: Topics Covered

🤖 Selection Process:
❄️ First Round Filtering: Applicants will undergo an initial screening process
❄️ Technical and Behavioral Interviews: Selected candidates from the first round will proceed to technical and behavioral interviews to assess their skills and suitability for the program

📋 Apply now to seize this opportunity and embark on a journey with A2SV!

🌟 Daily Algorithm Challenge! 🌟

Starting today, I’ll be posting a new question every day to help us prepare for A2SV in November! 💻✨

🗓 Challenge yourself and share your solutions!

Let’s sharpen our skills together. Ready to dive in?

Stay tuned for the first question! 🚀

👍 Like this post if you're excited for the daily algorithm challenges!

Your support motivates us to keep pushing our limits and learning together. Let’s make this journey fun and productive! 🚀

the answer is 042

Topic of This Week Linked List

we will work on this algorithm for the coming 7 days.

What is linked list https://youtu.be/XGSoQ0J5An4?si=-FmTaU2tI0LNP2tg

it is really a good podcast with a graduate from AAU in Information system degree

21. Merge Two Sorted Lists (Easy)

Problem:

You are given the heads of two sorted linked lists, list1 and list2.

Your task is to merge these two lists into one sorted linked list. The merged list should be formed by splicing together the nodes of list1 and list2.

Return: The head of the newly merged sorted linked list.

Examples:

Example 1:

Input: list1 = [1, 2, 4], list2 = [1, 3, 4]
Output: [1, 1, 2, 3, 4, 4]
Example 2:

Input: list1 = [], list2 = []
Output: []
Example 3:

Input: list1 = [], list2 = [0]
Output: [0]

try this question

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Find Middle of the Linked List
#Q19
Given a Singly Linked List, the task is to find the middle of the linked list. If the number of nodes are even, then there would be two middle nodes, so return the second middle node.

Example:

Input: linked list = 1 -> 2 -> 3 -> 4 -> 5
Output: 3
Explanation: There are 5 nodes in the linked list and there is one middle node whose value is 3.

Maximise the number of toys that can be purchased with amount K

Given an array consisting of the cost of toys. Given an integer K depicting the amount of money available to purchase toys. Write a program to find the maximum number of toys one can buy with the amount K.

Note: One can buy only 1 quantity of a particular toy.

Examples:

Input: N = 10, K = 50, cost = { 1, 12, 5, 111, 200, 1000, 10, 9, 12, 15 }
Output: 6
Explanation: Toys with amount 1, 5, 9, 10, 12, and 12 can be purchased resulting in a total amount of 49. Hence, maximum number of toys is 6.

how many of u can solve this problem with in 45min. share ur answer

answer:

def max_toys(cost, K):
    cost.sort()  
    count = 0
    total_cost = 0
    
    for price in cost:
        if total_cost + price <= K:
            total_cost += price
            count += 1
        else:
            break
            
    return count

N = 10
K = 50
cost = [1, 12, 5, 111, 200, 1000, 10, 9, 12, 15]
print(max_toys(cost, K))  # Output: 6