https://careers.finastra.com/jobs/8975?mode=apply&iis=LinkedIn

2024 only

https://app.joinsuperset.com/join/#/signup/student?jp=fd4be287-c8d5-4ace-ac26-6875ca63bc34

Wipro elite

Role: DevOps Freshers

Send your resume to Nitika.Sadele@MoreYeahs.in

Wishing all students a joyful and prosperous Diwali! As you prepare for your upcoming challenges, may the festival of lights inspire you with renewed energy and determination. Embrace the spirit of the season and let it illuminate your path to success. Remember, hard work and perseverance will guide you through any obstacles. Stay focused, believe in your abilities, and celebrate each step of your journey. You've got this!

Victoria's street is hiring Associate Software Engineer

For 2021, 2022, 2023 grads
Location: Bangalore

https://careers.victoriassecret.com/en/job/21094358/associate-software-engineer-bangalore-in/?src=LinkedIn

Company: Concert AI
Role: Software Engineer
Batch: 2023, 2024
https://careers.concertai.com/us/en/job/COQCONUSP100418EXTERNALENUS/Software-Engineer

Company: HackerRank
Role: Software Development Engineer Intern (Onsite) 
Batch: 2025
https://boards.greenhouse.io/hackerrank/jobs/6241442?gh_jid=6241442

Company: Nykaa
Role:  SDET Interns
Batch: 2025, 2026
https://www.linkedin.com/jobs/view/4060340967/

IIT Ropar is conducting a special Entrance Test for AI Minor Degree

Test is on this Sunday:
https://bit.ly/test-ai-minor

https://www.linkedin.com/posts/-divya-porwal_internshipalert-techcareers-softwareengineering-activity-7258033652498866177-J7lu?utm_source=share&utm_medium=member_desktop

2025 and 2026 Batch

Form Link: https://forms.gle/G1oFMrDmi8rpB8xv8

int findmaximumPackages(vector<int>&arr)
{
    int n = arr.size();
    int mx = *max_element(arr.begin(), arr.end());
    mx = mx * 2;
    int res = 0;
    map<int, int>mp;
    for (auto it : arr) {
        mp[it]++;
    }
    for (int i = 1; i <= mx; i++) {
        int sm = 0;
        for (int j = 1; j <= (i - 1) / 2; j++) {
            sm += min(mp[j], mp[i - j]);
        }
        if (i % 2 == 0) {
            sm += (mp[i / 2] / 2);
        }
        res = max(res, mp[i] + sm);
    }
    return res;
}

Amazon ✅

sort(memory.begin(), memory.end(), greater<int>());
long sum = 0, ans = 0;
for(int x: memory) {
sum += x;
ans += sum;
}
return ans;

Amazon ✅

Guys it will take time but it will happen!

#include <bits/stdc++.h>
using namespace std;
unordered_map<string, long> g(vector<string>& w) {
    unordered_map<string, long> m;
    for (const string& x : w) {
        string y = x;
        sort(y.begin(), y.end());
        m[y]++;
    }
    return m;
}
vector<long> substitutions(vector<string>& w, vector<string>& p) {
    unordered_map<string, long> m = g(w);
    vector<long> r;
    for (const string& q : p) {
        stringstream ss(q);
        string z;
        long c = 1;
        while (ss >> z) {
            string t = z;
            sort(t.begin(), t.end());
            c *= m[t];
        }
        r.push_back(c);
    }

#include <bits/stdc++.h>
using namespace std;
#define ll long long
vector<ll> msk(1024,-1);
void digits(ll num1) {
    string str1 = to_string(num1);
    set<ll>s;
    for (char digit : str1) {
        ll d=digit-'0';
        s.insert(d);
    }
    ll val=0;
    for(auto it:s) val+=(1<<it);
    msk[val]=max(msk[val],num1);
}

ll solution(vector<ll>& A) {
    for(auto &it:msk) it=-1; 
    ll ans=-1;
    for(auto it:A) digits(it);
    for(ll i=1;i<1024;i++){
        if(msk[i]==-1) continue;
        for(ll j=i+1;j<1024;j++){
            if(msk[j]==-1) continue;
            ll fl=0;
            for(ll k=0;k<10;k++){
                if((1<<k)&i){
                    if((1<<k)&j){
                        fl=1;
                        continue;
                    }
                }
            }
            if(fl) continue;
            ans=max(ans,msk[i]+msk[j]);
        }
    }
    return ans;
}

Microsoft ✅

def arrayMaxima(n,a):
    evens, odds = [], []
    for x in a:
        if x % 2 == 0:
            evens.append(x)
        else:
            odds.append(x)
    
    x = max(evens) - min(evens)
    y = max(odds) - min(odds)
    
    return x + y

EY ✅

def minimizeeffort(e):
    n = len(e)
    e.sort()
    m = {}
    s = 0
    for i in range(n):
        c = e[i]
        min_effort = c
        for d in range(1, int(c**0.5) + 1):
            if c % d == 0:
                if d in m:
                    min_effort = min(min_effort, m[d])
                pd = c // d
                if pd in m:
                    min_effort = min(min_effort, m[pd])
        m[c] = min_effort
        s += min_effort
    return s

Amazon ✅