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Leetcode with dani
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Forwarded from Codeforces Official
Codeforces Round #1012 (Div. 1, Div. 2) will take place on the 23rd of March at 05:35 UTC.
Please, join by the link https://codeforces.com/contests/2089,2090?locale=en
2169. Count Operations to Obtain Zero

Problem

You are given two non-negative integers, num1 and num2. In one operation, do the following:

• If num1 >= num2, subtract num2 from num1.

• If num1 < num2, subtract num1 from num2.

Repeat this until either num1 or num2 becomes zero. Return the total number of operations performed.

Examples

Example 1

Input:
num1 = 2, num2 = 3
Output:
3

Example 2

Input:
num1 = 10, num2 = 10
Output:
1

Constraints

• 0 ≤ num1, num2 ≤ 10⁵
Leetcode with dani
▎2169. Count Operations to Obtain Zero ▎Problem You are given two non-negative integers, num1 and num2. In one operation, do the following: • If num1 >= num2, subtract num2 from num1. • If num1 < num2, subtract num1 from num2. Repeat this until either…
class Solution:
def countOperations(self, num1: int, num2: int) -> int:
count = 0
while num1 and num2:
if num1>=num2:
count += num1//num2
num1 = num1%num2
else:
count += num2//num1
num2 = num2%num1
return count
can solve this question using recursion ?
🔵 1823. Find the Winner of the Circular Game

📌 Problem:

There are n friends sitting in a circle, numbered 1 to n.

Starting from friend 1, count k friends clockwise (including the current one).

The k-th friend leaves the circle.

Repeat the process, starting from the next friend.

The last remaining friend is the winner.

📌 Input:

n → Number of friends.

k → Step count for elimination.

📌 Output:

The winner's number.

📌 Example 1:
🔹 Input: n = 5, k = 2
🔹 Output: 3
🔹 Explanation:
Friends leave in this order: 2 → 4 → 1 → 5 → (Winner: 3)

📌 Example 2:
🔹 Input: n = 6, k = 5
🔹 Output: 1
🔹 Explanation:
Friends leave in this order: 5 → 4 → 6 → 2 → 3 → (Winner: 1)

🔎 Can you find the last friend standing? 😃
In-person Conversion Registration is Open! 🚀

We’re excited to welcome students for the A2SV in-person conversion! This is your chance to take your learning to the next level and be part of a vibrant tech community.

Open to current students from Addis Ababa University (AAU), Addis Ababa Science and Technology University (AASTU), and Adama Science and Technology University (ASTU).

📅 Registration Dates: March 24 - March 28

Requirements:
- At least 150 solved problems
- At least 30 active days

📚 Topics to Cover: Two Pointers, Sorting, Sliding Window, Stack, Queue, Monotonicity, Linked List, Recursion.

Don’t miss this opportunity—apply now!

🔗 Apply here: link

#A2SV #TechEducation #CodingJourney #LevelUp
👍5
Eid Mubarak! Wishing you a day full of joy and blessings!
8🥰1
Forwarded from Codeforces Official
🤪April Fools Day Contest 2025🤪 will take place on the 1st of April at 14:35 UTC.
Please, join by the link https://codeforces.com/contests/2095
If anyone has questions or is interviewed by A2SV today, please share the questions with me. I'll post them in this channel!
@zprogramming_bot
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Question Denoscription
You are given a 0-indexed string s consisting of only lowercase English letters. Return the number of substrings in s that begin and end with the same character.
A substring is a contiguous non-empty sequence of characters within a string.

Example 1:
Input: s = "abcba" #"a", "b", "c", "b","a",abcba","bcb" a:2 b:2 c : 1
Output: 7
Explanation:
The substrings of length 1 that start and end with the same letter are: "a", "b", "c", "b", and "a".
The substring of length 3 that starts and ends with the same letter is: "bcb".
The substring of length 5 that starts and ends with the same letter is: "abcba".

Example 2:

Input: s = "abacad"
Output: 9
Explanation:
The substrings of length 1 that start and end with the same letter are: "a", "b", "a", "c", "a", and "d".
The substrings of length 3 that start and end with the same letter are: "aba" and "aca".
The substring of length 5 that starts and ends with the same letter is: "abaca".

Example 3:

Input: s = "a"
Output: 1
Explanation:
The substring of length 1 that starts and ends with the same letter is: "a".


Constraints:

1 <= s.length <= 105
s consists only of lowercase English letters
Forwarded from Ibnu-Nezif Abasambi
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);
}
41
Forwarded from Dagmawi Babi
Was talking to Emre and yeah A2SV is staying and the issues are resolving. There's still an issue of funding but he's very hopeful about it now. We won :)

I'm super glad and I'm happy for all the present and future A2SVians. This's wonderful news. Thank you everyone that supported, shared and encouraged this. ❤️
👍9🤗3
When someone mentions on LinkedIn "I finally landed my dream job", they miss to tell you about

Let's think about it step by step.

Why Do You Need a Job?

Before jumping into any career path, ask yourself:

➤ Why do I need a job?

Is it for financial stability?

Personal growth?

A sense of purpose?

Jobs can provide structure, income, and learning opportunities. But they are not the only way to achieve these goals. Clarity on your "why" will help you make smarter decisions.

What if the market is not in your favor, which it is right now?

The job market is unpredictable. Recessions, layoffs, and industry shifts happen. So, what's your backup plan?

Here are some alternatives:

Freelancing: Offer your skills to clients worldwide. Platforms like Upwork or Fiverr make this easier than ever.
Starting a Side Hustle: Turn hobbies into income. Think tutoring, content creation, or selling digital products.

Remote Work: Explore temporary or part-time remote jobs to stay flexible while earning.

Your job shouldn't be your only source of income. Think of ways to diversify.

How Can You Acquire More Skills to Turn the Tables?

If the market isn't working for you, invest in yourself.

➤ Here's how:

Take Online Courses: Platforms like Coursera or Udemy offer affordable courses in trending fields.

Network: Attend webinars, join LinkedIn groups, and connect with industry leaders.

Experiment: Take on small projects outside your comfort zone to build confidence and expertise.

Every new skill adds value to your profile and opens new opportunities.

Look for jobs that offer more than a paycheck

A great job isn't just about money.

➤ It's about:

Learning: Does the role teach you something new?

Growth: Will it help you progress in your career?

Stability: Does it offer financial and mental security?

If a job doesn't check these boxes, it might be time to reassess.
Hustle, but avoid toxicity

➤ Hard work is essential, but remember:

Hustle is healthy; toxicity is not.

Your mental health and well-being are non-negotiable.

Walk away from environments that drain you. Success should feel empowering, not exhausting.

So,

Build multiple income streams.

Keep growing and learning.

Prioritize your happiness and health.

So, what's your next step? Are you building skills, exploring alternatives, or rethinking your path? Tell me in the comments below!
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