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AccioJob Offline Hiring Drive with Intellinez Systems.

Role: Trainee Engineer (After the Internship, the Position will be a Software Engineer.)
โ Graduation Year: 2023,2024,2025
โ 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
โ Work Location: Hyderabad
โ Internship Stipend - 25K per month for 6 months
CTC: โ‚น5 to โ‚น5.5 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.
๐Ÿ‘2
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;
}

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