AccioJob Offline Hiring Drive with Intellinez Systems.

Role: Trainee Engineer (After the Internship, the Position will be a Software Engineer.)
⁠Graduation Year: 2023,2024,2025
⁠Work location: During Internship - Mohali and after full-time -Bangalore
CTC - 4 LPA, Internship - Stipend - 15K for 6 Months
Register here: https://lnkd.in/g5wP4gXJ


Note: Bond:- 2 Years, applicable once you become full-time. The penalty is 1.5 Lac.

Evaluation Process
Date: 13th, 14th and 15th Jan in Noida Only
Offline Assessment at AccioJob Noida Skill Centre
Company Process - 2 technical interviews> 1 HR Round

Register here: https://lnkd.in/g5wP4gXJ

Important: Bring your laptop for the test.

AccioJob Offline Hiring Drive with OSI Systems for Java Developer & Intern

Graduation Year: 2022, 2023, 2024 & 2025
⁠Work Location: Hyderabad
⁠Internship Stipend - 25K per month for 6 months
CTC: ₹5 to ₹5.5 LPA
Register here: https://lnkd.in/gYVy-bii


Evaluation Process
Date: 12th, 17th & 18th Jan in Hyderabad only
Offline Assessment at AccioJob Hyderabad Skill Centre
Company Process - 2 technical interviews> 1 HR Round

Register here: https://lnkd.in/gYVy-bii

Important: Bring your laptop for the test.

Company Name: Hyland

Role: Developer 1

https://careers-hyland.icims.com/jobs/12000/developer-1/job

Company Name: Flipkart
Competition: Flipkart Girls Wanna Code
Eligibility: 2025 & 2026 female passouts
Internship Stipend: INR 1 Lacs per month
SDE-1 Role Compensation: INR 32.67 LPA

Apply link :

https://unstop.com/hackathons/flipkart-girls-wanna-code-60-flipkart-1349594

Do share this with your Friends too

📌Diebold Nixdorf is hiring Associate Software Engineer

Experience: 0-2 yrs

Expected Salary: 6-12 LPA

💻Apply Link: https://eeug.fa.us6.oraclecloud.com/hcmUI/CandidateExperience/en/sites/CX/job/14293

https://www.linkedin.com/posts/mitrahsoft_softwaredevelopers-softwareengineers-maduraijobs-activity-7283447614811860992-5H5s/

Tamil Candidates are preferred

https://deloittehacksplosion.hirepro.in/#details

2025 and 2026 batch

https://www.linkedin.com/posts/shardul-deshpande-01_qualcomm-internship-summer2025-activity-7284024100916809730-O3-6?utm_source=share&utm_medium=member_desktop

Summer 2025 Internship Opportunity 🚀

The Chipset Power Team at Qualcomm is hiring! We’re looking for two MS or PhD students in CS, CE, or EE (graduating by 2026) with a strong understanding of computer architecture and working knowledge of Python.

Join us to work on cutting-edge technologies and make an impact in the semiconductor industry. If this sounds like you, or someone you know, reach out or share this post!

https://cutshort.io/job/Software-Developer-Intern-Grad-2025-Pune-Hummingbird-Web-Solutions-Private-Limited-iWNQ486r?utm_source=linkedin-feed&utm_medium=xml&utm_content=job-posting&applicationsource=linkedin-feed

2025 Batch

https://job-boards.greenhouse.io/episodesixlinkedin/jobs/5411834004?gh_src=d1178d084us

2024 and 2025

Zeta is hiring for SDE Intern (2025 grads)

Apply: https://jobs.lever.co/zeta/7703ac43-7425-4af3-8551-9f53eeb9bc6e?lever-via=brPz4nskIn&lever-social=job_site

Honeywell Hiring !!
Role - C++ Developer
Exp - fresher

Link - https://careers.honeywell.com/us/en/job/HONEUSREQ475604EXTERNALENUS/C-Developer

BT Group is hiring for Trainee Associate Engineer

Experience: 0 - 1 year's
Expected Salary: 4-6 LPA

Apply here: https://jobs.bt.com/BTGroup/job/Building-No-14-Sector-24-&-25A-Trainee-Associate-Engineer-MID-Guru/811114202/

typedef long long ll;

ll solution(int n, vector<int> a1, vector<int> a2) {
    vector<ll> left(n+1, 0), right(n+2, 0), prefixSum(n+1, 0), maxL(n+1, 0), maxM(n+1, LLONG_MIN);
    for (int i = 1; i <= n; i++) {
        left[i] = i == 1 ? a1[0] : max(left[i-1] + (ll)a1[i-1], (ll)a1[i-1]);
    }
    ll res = LLONG_MIN;
    for (int i = 1; i <= n; i++) {
        res = max(res, left[i]);
    }
    for (int i = n; i >= 1; i--) {
        right[i] = i == n ? (ll)a1[n-1] : max(right[i+1] + (ll)a1[i-1], (ll)a1[i-1]);
    }
    for (int i = 1; i <= n; i++) {
        prefixSum[i] = prefixSum[i-1] + (ll)a2[i-1];
    }
    maxL[1] = left[0] - prefixSum[0];
    maxM[1] = maxL[1];
    for (int i = 2; i <= n; i++) {
        maxL[i] = left[i-1] - prefixSum[i-1];
        maxM[i] = max(maxM[i-1], maxL[i]);
    }
    for (int j = 1; j <= n; j++) {
        res = max(res, maxM[j] + prefixSum[j] + right[j+1]);
    }
    return res;
}

Uber She ++✅