Employee information
Nextwave ✅

import java.util.Scanner;

public class Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int energy = sc.nextInt();
        int maxDrink = sc.nextInt();
        int maxGoodEnergy = 0;
        int maxGoodNumber = 0;
        for (int i = 1; i <= maxDrink; i++) {
            int goodEnergy = energy * i;
            int goodNumber = 0;
            while (goodEnergy % 10 == 0) {
                goodNumber++;
                goodEnergy /= 10;
            }
            if (goodNumber > maxGoodNumber) {
                maxGoodNumber = goodNumber;
                maxGoodEnergy = energy * i;
            } else if (goodNumber == maxGoodNumber) {
                maxGoodEnergy = Math.max(maxGoodEnergy, energy * i);
            }
        }
        System.out.println(maxGoodEnergy);
    }
}

Airtel shecode ✅

#include<bits/stdc++.h>
using namespace std;

bool solve(char c) {
    c = tolower(c);
    return !(c == 'a' c == 'e' c == 'i' c == 'o' c == 'u') && c >= 'a' && c <= 'z';
}

int main() {
    string s;
    cin >> s;
    int count = 0;
    for(int i = 0; i < s.length(); i += 2) {
        if(solve(s[i])) {
            count++;
        }
    }
    cout << count << endl;
    return 0;
}

Airtel Shecode ✅
anabellle

#include<bits/stdc++.h>
using namespace std;

int main() {
    string s;
    cin >> s;
    int n = s.size();
    vector<int> prefix_b(n, 0);
    prefix_b[0] = (s[0] == 'b');
    for(int i = 1; i < n; i++) {
        prefix_b[i] = prefix_b[i-1] + (s[i] == 'b');
    }
    int ans = prefix_b[n-1];
    for(int i = 0; i < n; i++) {
        if(s[i] == 'a') {
            ans = min(ans, prefix_b[i] + (n - i - 1 - (prefix_b[n-1] - prefix_b[i])));
        }
    }
    cout << ans << endl;
    return 0;
}

Converter
Airtel Shecodes✅

#include <iostream>

using namespace std;

int main() {
    int e, d;
    cin >> e >> d;
   
    int mE = 0;
    int mN = 0;
   
    for (int i = 1; i <= d; i++) {
        int gE = e * i;
        int gN = 0;
       
        while (gE % 10 == 0) {
            gN++;
            gE /= 10;
        }
       
        if (gN > mN) {
            mN = gN;
            mE = e * i;
        } else if (gN == mN) {
            mE = max(mE, e * i);
        }
    }
   
    cout << mE << endl;
   
    return 0;
}


shecode goodenergy ✅

#include <bits/stdc++.h>
using namespace std;

#define ll long long

int main() {
    ll n;
    cin >> n;
    vector<ll> v(n);
    for(ll i=0; i<n; i++)
        cin >> v[i];

    sort(v.begin(), v.end());

    ll sum = 0;
    for(ll i=0; i<n; i++) {
        if(sum + v[i] < 0)
            return cout << -1, 0;
        sum += v[i];
    }

    if(sum % 2 == 0)
        cout << -1;
    else
        cout << sum;

    return 0;
}


Shecode fortunejob✅

Hiring for business analyst (freshers)

https://jpmc.fa.oraclecloud.com/hcmUI/CandidateExperience/en/sites/CX_1001/job/210509109

import heapq

def distance(x, y):
    return x*x + y*y

def nearest_houses(P, T, queries):
    distances = []
    heapq.heapify(distances)

    for query in queries:
        if query[0] == 1:
            x, y = query[1], query[2]
            heapq.heappush(distances, distance(x, y))
        else:
            nearest = heapq.nsmallest(T, distances)[-1]
            print(nearest)

Nearest House ✅

def max_call_executives(n, start_times, end_times):
    timeline = [0] * (24 * 60 + 1)
   
    for i in range(n):
        start = int(start_times[i][:2]) * 60 + int(start_times[i][2:])
        end = int(end_times[i][:2]) * 60 + int(end_times[i][2:])
        timeline[start] += 1
        timeline[end] -= 1
   
    max_executives = 0
    current_executives = 0
    for i in range(len(timeline)):
        current_executives += timeline[i]
        max_executives = max(max_executives, current_executives)
   
    return max_executives

Call Centre ✅

Priyanka:
Jp morgan interview completed..these r the two questions..How communications plays a mojor role in the job environment...and 2nd one How will u tell about the topic that you u know very well to others.

There will 2 tries..u can do retries..so dont wry about 1st attempt.

Salesperson ID's - IBM✅

Lexicographical smallest substring - IBM ✅

Summer Internship positions for students at Bosch Corporate Research, India.
 
Domains: Text Analytics/NLP/Generative AI.
Duration: 2-6 months or more.
Responsibilities: Research activities on Natural Language Processing, Text Analytics, and Deep Learning
- Large Language Models
- RAG
- Domain Adaptation
- Knowledge Graphs
 
Qualifications: Current Bachelor / Master / Ph.D. students of Computer Science. Strong knowledge of Machine Learning, Deep Learning, and NLP. Interest and aptitude in research activities.
 
Interested candidates can send resumes at: 
- rtc_careers@in.bosch.com
- Subject: CR/RTC-IN: LLMs Internship - <name>, <college>, <degree_program (B.Tech/M.Tech./M.S/Ph.D.)>, <branch>, <graduation_year>
- e.g. CR/RTC-IN: LLMs Internship - Kartheek, IIT-H, M.Tech, CSE, 2025

#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll count(ll k,vector<ll>&a,vector<ll>&b)
{
    ll n=a.size();
    ll m=b.size();
    unordered_map<ll,ll>mpp;
    ll ans=0;
    for(ll i=0;i<n;i++) mpp[__gcd(a[i],k)]++;
    for(ll i=0;i<m;i++)
    {
        for (auto &it:mpp)
            if((__gcd(b[i],k)*it.first)%k==0) ans+=it.second;
    }
        return ans;
}
signed main() 
{
    ll n,m,k; cin>>n>>m>>k;
    vector<ll>a(n),b(m);
    for(ll i=0;i<n;i++) cin>>a[i];
    for(ll i=0;i<m;i++) cin>>b[i];
    cout<<count(k,a,b);
    return 0;
}

Product divisibility ✅

Remote Intern Hiring

Hiring Process, Work at Rapidfort
https://youtu.be/aNBX-KbkWlE?si=KV9m_ogYy9KSIkFa