Enjoy but don't get attached Give but dont expect Listen but dont judge Love but not depend and finally accept but dont settle

Are you an African who dreams of building Google-level software engineering skills?

Applications are now open for A2SV G7 Education!

This is your moment to learn, grow, and prepare for the top tech companies you’ve always aspired to join.
At A2SV Education, you’ll join a vibrant community that challenges you to think bigger, code smarter, and create real impact across Africa 🌍.

⏱️ It only takes 7 minutes to apply. Take your shot at joining A2SV G7 Education today!

👉 Apply now: https://form.typeform.com/to/wOWdCFrZ

📣 Spread the word! Tag a friend, share this post, and help us reach more brilliant African minds ready to transform their future through A2SV.
Together, we empower Africa’s next generation of world-class engineers and tech entrepreneurs! 💪 💻

#A2SV #Education #Generation7 #TopTech #ApplicationsAreOpen #SoftwareEngineering #AfricaTech

can u solve this puzzel

📝 Problem: Dividing Students

A teacher wants to divide a group of students into 3 subgroups for practice. Each student has a favorite programming language:

a students like Assembler

b students like Basic

c students like C++

The teacher knows that Assembler fans and C++ fans argue if they are in the same subgroup, so no subgroup should contain both Assembler and C++ fans.

The goal is to divide the students into 3 subgroups so that the largest subgroup is as small as possible.

Input

First line: t — number of test cases (1 ≤ t ≤ 5)

Each test case: three integers a b c (1 ≤ a, b, c ≤ 1000)

Output

For each test case: print the minimum possible size of the largest subgroup\

📣 A2SV Education — Generation 7 Update!

Hey A2SV Community 👋,

Here are a few important clarifications about the Generation 7 Education Program👇

🔹 1️⃣ In-Person Education (AAU)
The education will continue in person for Addis Ababa University (AAU) students.
📍Note: Sessions might not take place inside one of the AAU campuses, but they will be held nearby within walking distance.

🔹 2️⃣ Eligibility for Continuing Students
If you’re currently studying at AAU, AASTU, or ASTU, you’re only eligible for the in-person program.
❌ The remote option is not available for these students.

🔹 3️⃣ Upcoming Public Sessions
We’ll soon hold public info sessions to:
✨ Answer your most common questions
✨ Introduce A2SV Education more deeply
✨ Explain key prerequisites before the interviews

Stay tuned for the schedule! 📅

⭐️ Spread the Word!
Please share the application form with other students who might benefit from this opportunity.
Let’s help more people learn, grow and succeed through A2SV Education 🚀

— A2SV Education Team

Last year interview

Report on my Interview

Literally when I started my PC stacked🤯 and he was a peace guy he literally gave me 20 min after that I changed my PC with my friend and shared that.
1. After I shared my PC he introduced himself to me and also told me to share my screen asked me to go from tab desktop to check whether I shared my screen.
Also he asked me to show my ID in front of the screen.

2. He dropped the question it was

leetcode 1198 Premium question (https://github.com/doocs/leetcode/blob/main/solution/1100-1199/1198.Find%20Smallest%20Common%20Element%20in%20All%20Rows/README_EN.md)

3. I asked him to give me some time to understand the problem. When I saw the word strictly increasing I thought the question was Monotonic Stack but it wasn't.
4. After few silent minutes I think my adrelanine hormone just increased and the optimal one came to me but not the brute.
5. So I thought of a way to just sort the dictionary and make explain it as brute force. And literally it worked and asked me to Optimize it after I explained the time complexity. He also guided me on some other optimal related things like he told me to finish it inside the nested loop since it is strictly increasing matrix so I literally did that and explained as Optimal form and after I explained I implemented it quickly I started implementing while I was explaining the Brute one.
6. The technical was like that we finished it in 30min after that he gave 3 min and copy the code run and submit. He didn't told me whether it was accepted but I personally tested it on sharepad and it was correct for th given testcases.
7. After 5 min we started the Behavioural Interview he asked me 7 questions and agree on the policy the questions was literally same as the

G4 Interview Template

so I highly recommend to prepare on it.
8. After the interview I asked him about A2SV and other tech staff since he was Graduate student. And the interview entirely ended in 1:10 hr.

I will share interview template and more interview questions

Good Morning, A2SV community! 🌞
Here’s a complete guide to help you navigate the G7 remote education journey 🎓, understand the interview process 💬, and meet the required prerequisites ✅.

G6 Interview Question

I HAVE INTERVIEWED ON A2SV ON THURSDAY AND THE QUESTION WAS



/*
Question Denoscription
Given a string s and an integer k, return the number of substrings in s of length k with no repeated characters.

Example 1:

Input: s = "havefunonleetcode", k = 5
Output: 6
Explanation: There are 6 substrings they are: 'havef','avefu','vefun','efuno','etcod','tcode'.

Example 2:

Input: s = "home", k = 5
Output: 0
Explanation: Notice k can be larger than the length of s. In this case, it is not possible to find any substring.


Constraints:
1 <= s.length <= 104
s consists of lowercase English letters.
1 <= k <= 104
*/


FINALLY I HAVE SOLVED THE QUESTION LIKE THIS IN C++ MAY BE IT IS USEFUL

#include <bits/stdc++.h>
using namespace std;
class Solution {
    public :
        int solve (string s,int k){
            vector <int> freq(26);
            
            int l=0;int r=0;
            int offset=(int)'a';
            int ans=0;
            while (r<s.size()){
                if (r-l<k-1){   
                   freq[s[r]-offset]++;
                    r++;             
                }else if (r-l==k-1){
                    freq[s[r]-offset]++;
                    r++;
                    bool t=true;
                    for (int x:freq){
                        if (x>1){
                          t=false;
                        }
                    }
                    t?ans+=1:ans=ans;
                    
                }else if (r-l>k-1){
                    freq[s[l]-offset]--;
                    l++;
                }
                
            }
            return ans;
        }
};

int main () {
    string s="abcdefhijklmnopqrstuvwxyz";
    int k=5;
    Solution* sol=new Solution();
    cout<<sol->solve(s,k);
}

Last year interview

Todays another interview questions