🚨 Exciting News 🚨
ALX Ethiopia is NOW OPEN! 🎉

Ready to kickstart your career with world-class tech training? 🚀
💰 Only $5/month ✔️
❌ $200 💸

Here are the amazing programs available to you:

🔹 ALX Pathway
🔹 Virtual Assistant
🔹 Professional Foundations
🔹 AI Career Essentials
🔹 Front-End Web Development
🔹 Back-End Web Development
🔹 AWS Cloud Computing
🔹 Salesforce Administrator
🔹 Data Science
🔹 Data Analytics

Don't miss out— Get started for only $5/month! 💥
👉 Learn More & Apply: ALX Programs
👉 Get the full ALX Entrance Assessment Answers: ALX Assessment key

#TechTraining #CareerGrowth #ALX #Ethiopia #FutureLeaders 🌍✨

🌟🚀 @AceCoding Presents! 🚀🌟

🌟 A2SV Past Interview Questions 🌟


Hey everyone! 📢 If you're preparing for an A2SV interview, here are some
common questions you might encounter. Get ready to shine! ✨

Common Questions:

1.Tell me about yourself.
2. What do you know about A2SV and why do you want to join us?
3. Describe a time when you had to step out of your comfort zone to achieve something.
4. How do you give back to your community?
5. What are your strengths and weaknesses?

Additional Questions:


Introduce yourself 👩.
What is the hardest decision you've had to make?
Tell us about a time when there was a disagreement in a group. How did you resolve it?
How did you hear about A2SV?
Are you committed to the cause?


Good luck with your preparations! You've got this. 💪😊

Feel free to reach out if you need more tips or have any questions. Let's conquer this together! 🚀👩‍💻👨‍💻



I hope this will makes you feel more prepared and confident! If there's anything more you'd like to add or adjust, just let me know in the discussion group 😊.

🟡 Medium leetcode question: 287. Find the Duplicate Number

🟢 Using Cycle sort

class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        i = 0 
        while i < len(nums):
            correct_pos = nums[i] - 1
            if nums[i] != nums[correct_pos]:
                nums[i], nums[correct_pos] = nums[correct_pos], nums[i]
            else:
                i +=1 
        
        for i in range(len(nums)):
            if nums[i] != i + 1:
                return nums[i]

👨‍💻 Optimized Cycle sort

class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        i = 0 
        while i < len(nums):
            if nums[i] != i + 1:
                correct_pos = nums[i] - 1
                if nums[i] != nums[correct_pos]:
                    nums[i], nums[correct_pos] = nums[correct_pos], nums[i]
                else:
                    return nums[i]
                
            else:
                i += 1
        
        return -1

🔴 Cycle detection Algorithm (Floyd's torties and hare / slow and fast pointer algorithm)

class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        slow = nums[nums[0]]
        fast = nums[nums[nums[0]]]

        while slow != fast:
            slow = nums[slow]
            fast = nums[nums[fast]]
        
        slow = nums[0]

        while slow != fast:
            slow = nums[slow]
            fast = nums[fast]
        
        return slow

Time complexity : O(n) & Space Complexity: O(1)

🌟🚀 @AceCoding Presents! 🚀🌟

https://leetcode.com/problems/find-the-duplicate-number/

🔴 Cycle detection Algorithm (Floyd's torties and hare / slow and fast pointer algorithm)

class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        slow = nums[nums[0]]
        fast = nums[nums[nums[0]]]

        while slow != fast:
            slow = nums[slow]
            fast = nums[nums[fast]]
        
        slow = nums[0]

        while slow != fast:
            slow = nums[slow]
            fast = nums[fast]
        
        return slow

🌟🚀 @AceCoding Presents! 🚀🌟

🟡 1561. Maximum Number of Coins You Can Get

💻 This is a very good question on Greedy Algorithm and sorting
for those who don't know Greedy is a type of Algo or technique used to solve some problems efficiently ( By being GREEDY every time 😄 that is why it's called greedy)

📊 Most Efficient Solution both in time and space complexity

class Solution:
    def maxCoins(self, piles: List[int]) -> int:
        piles_per_person = len(piles) // 3
        piles.sort(reverse=True)
        sum = 0 
        for i in range(1, len(piles) - piles_per_person, 2):
            sum += piles[i]
        return sum

Time complexity : O(n*log n) & Space Complexity: O(1)

🌟🚀 @AceCoding Presents! 🚀🌟

https://leetcode.com/problems/maximum-number-of-coins-you-can-get/

👨‍💻 Here is a detail explanation for the above question👆

class Solution:
    def maxCoins(self, piles: List[int]) -> int:
        # first determine how many piles each person will get 
        piles_per_person = len(piles) // 3
        # sort the piles in descending order 
        piles.sort(reverse=True)
        # Hypothetically we want to give the least piles in the sorted pile for bob
        # [9,8,7,6,5,4,3,2,1] 1, 2, 3 are given for bob
        p = piles_per_person 
        score = piles[1:-p:2] # starting from the second value and increment by 2 until we reach len(piles) - piles_per_person
        return sum(score)

🌟🚀 @AceCoding Presents! 🚀🌟

https://leetcode.com/problems/maximum-number-of-coins-you-can-get/

📢 A2SV past years interview questions 📝


©️Ethio Toolkit

You have some apples and a basket that can carry up to 5000 units of weight. Given an integer array weight where weight[i] is the weight of the ith apple, 

return the maximum number of apples you can put in the basket.

Example 1:

Input: weight = [100,200,150,1000]

Output: 4

Explanation: All 4 apples can be carried by the basket since their sum of weights is 1450.

Example 2:

Input: weight = [900,950,800,1000,700,800]

Output: 5


Constraints:

1 <= weight.length <= 1000

1 <= weight[i] <= 1000

This problem is lighter version of a classic interview question: "1561. Maximum Number of Coins You Can Get" 💰which we have solved on Ace Coding.

✅ Solution: ©️Ace Coding

def maxNumberOfApples(weight):
    weight.sort()
    total_weight = 0
    count = 0

    for w in weight:
        if total_weight + w <= 5000:
            total_weight += w
            count += 1
        else:
            
    return count

🚀 @AceCoding Presents! 🚀🌟

🔔 Quick Reminder!

If you found this problem helpful, make sure to share it with your friends. Remember the saying:

If you want to go fast, go alone. If you want to go far, go together. 💪

Let’s solve these challenges together and grow as a community! 🚀

🌟🚀 @AceCoding Presents! 🚀🌟

✅ A2SV G5 Camp Topics

🌟🚀 @AceCoding Presents! 🚀🌟

🟡 189. Rotate Array

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        rotation_point = k % len(nums)
        k = rotation_point
        if k != 0: 
            nums[:k], nums[k : ] = nums[-k : ], nums[:-k]
        # the rotation_point should be counted for the end of nums
        # Ex. [1, 2, 3, 4, 5, 6, 7]  nums[:k] = [1, 2, 3]
        # But nums[:-k] = [1, 2, 3, 4] which is correct
        # nums[-k:] = [5, 6, 7]

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        k = k % len(nums)
        
        def reverse(left, right): 
            while left < right:
                nums[left], nums[right] = nums[right], nums[left]
                right -= 1
                left += 1
            
        reverse(0, len(nums)-1)
        reverse(0, k-1)
        reverse(k, len(nums)-1)

🌟🚀 @AceCoding Presents! 🚀🌟

https://leetcode.com/problems/rotate-array/description/

🚀 Heads up, everyone! 🚀

The questions we share here will likely appear in the A2SV technical interviews. Keep an eye on them and practice regularly to boost your preparation! 💪🔍

Happy Coding💻

Stay tuned and don't miss out on these key problems. 🧠✨

📣 Update Alert: Info Session RSVP & Location Details!

🚀 Great News! We’ve expanded the RSVP tickets to accommodate more attendees! If you haven’t RSVP’d yet, now’s your chance to grab a spot and join the action.

👉 RSVP Now to secure your spot : GDG AASTU x AASTU SEA INFO SESSION 😎

📍 Location Reminder: The Info Session will take place at AASTU Old Graduation Hall.

📎 Study Session Registration:
Interested in the upcoming student study session? Registration will be open at the venue during the event. Be sure to attend to secure your spot!

Spread the word, and we can’t wait to see you on 20 November 2024! Let’s make this event unforgettable. ✨

🚀 Join Us for the INFO SESSION by AASTUSEA! 🚀

Ready to dive into tech? Whether you're brand new, dreaming of becoming a developer, or already coding away, this event is for YOU! No experience needed—just bring your passion and excitement!

✨ What's in Store?

💫 Connect with Google Developer Groups (GDG) and AASTUSEA

💫 Learn about plans and opportunities provided by the Association.

💫 Secure your spot in our project preregistration.

💫  Get the Details needed to ace Huawei ICT Competition 2024-25, Join Huawei ICT Academy and Get FREE certifications.

💫Learn the basics, get inspired, and kickstart your coding journey

💫Network with fellow tech enthusiasts and mentors

📅 Date: November 20, 2024 
🕗 Time: 8:30 - 11:00 LT 
📍 Location: Old Graduation Hall

👉👉 RSVP Here  👈👈

Don't miss this great opportunity to connect and collab with like-minded once!

@AASTUSEA aastu.software

🚀 Part 2 Quiz: Computer Organization and Architecture 🖥

We’re back with another round of quiz questions on Computer Organization and Architecture! 🧠

🌟🚀 @AceCoding Presents! 🚀🌟