Hi All
Qualcomm is looking for Mtech/Btech Fresh Grads 2023 passouts to work with our Software Dev/Test teams in Hyderabad/Bangalore locations. If interested please share your resume to aayushar@qti.qualcomm.com

*apply all batches and hope you will get accepted*

Hello #Connections
@ECI is conducting a #SOC #Walk-In drive on 12-Jan-2024 (Friday) in Bangalore Office.
Positions -
1. SOC Analyst (Experience - Freshers OR 0-1 year)
1. SOC Associate (Experience - Freshers OR 0-1 year)
Location - #Bangalore
#Skills - Cyber Security, SIEM, SOC Trainings, SOC Certifications

To book your interview slot, please apply at rrai@eci.com

NVIDIA is hiring through their new graduate hiring program named N.Ex.T(NVIDIA Exceptional Talent).

𝐏𝐫𝐨𝐜𝐞𝐬𝐬: 1. Two rounds of online technical tests via HackerRank, followed by two rounds of discussions for candidates who clear the tests.
2. Selected candidates will be offered a full-time New College Graduate (NCG) position in the Hardware Engineering group at NVIDIA in India.

𝐄𝐥𝐢𝐠𝐢𝐛𝐢𝐥𝐢𝐭𝐲 𝐂𝐫𝐢𝐭𝐞𝐫𝐢𝐚: 1. Students of B.Tech/ Dual degree/ M.Tech/ MS from CS/ EE/ EC Departments, graduating in 2024. 

2. CGPA of 6 and above. 𝐓𝐢𝐦𝐞𝐥𝐢𝐧𝐞: 1. Last Day to Apply: 2nd Feb, 2024 (11 PM) 2. N.Ex.T Basic Test - January 28th, 2024 (Analytical Reasoning, Problem Solving and Basic Technical Skills Test)

3. N.Ex.T Advanced Test – February 4th, 2024 (Hardware Technical Test)

https://www.hackerrank.com/event/next

Odd Even Code
Python 3✅
IBM

DNA code
IBM✅

Valid user ✅

vector<int>solve(vector<int> arr, int transform) {
    for(int t = 0; t < transform; t++) {
        for(int i = 0; i < arr.size(); i++) {
            if(arr[i] % 2 == 0) {
                arr[i] -= 3;
            } else {
                arr[i] += 3;
            }
        }
    }
    return arr;
}

Odd Even ✅

string solve(string bs) {
    map<string, string> nb = {
        {"001", "C"},
        {"010", "G"},
        {"011", "A"},
        {"101", "T"},
        {"110", "U"},
        {"000", "DNA"},
        {"111", "RNA"}
    };

    string ds = "";
    string t = nb[bs.substr(0, 3)];
    for(int i = 3; i < bs.length(); i += 3) {
        string b = bs.substr(i, 3);
        if(nb.find(b) != nb.end()) {
            string x = nb[b];
            if(t == "DNA" && x == "U") {
                x = "T";
            }
            ds += x;
        } else {
            ds += "Error";
        }
    }

    return ds;
}

DNA✅
IBM

bool isPal(int n) {
    int r, s = 0, t;
    t = n;
    while (n > 0) {
        r = n % 10;
        s = (s * 10) + r;
        n = n / 10;
    }
    return (t == s);
}

int firstPal(int n) {
    int i = 1;
    while (true) {
        if (isPal(i)) {
            int d = 1 + log10(i);
            if (d == n)
                return i;
        }
        i++;
    }
}

void login(int d, string u, string p) {
    map<string, string> users = {
        {"user1", "pass1"},
        {"user2", "pass2"},
        {"user3", "pass3"},
        {"user4", "pass4"},
        {"user5", "pass5"}
    };

    if (users.find(u) != users.end() && users[u] == p) {
        int t = firstPal(d);
        cout << "Welcome " << u << " and the generated token is: token-" << t << endl;
    } else {
        cout << "UserId or password is not valid, please try again." << endl;
    }
}

IBM✅

vector<int> solve(int n, vector<vector<int>> m) {
    vector<int> t(n, 0);
    vector<int> w(n, 0);

    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            if(m[i][j] == 1) {
                t[i]++;
            }
        }
    }

    int maxi = *max_element(t.begin(), t.end());

    for(int i = 0; i < n; i++) {
        if(t[i] + count(m[i].begin(), m[i].end(), 2) >= maxi) {
            w[i] = 1;
        }
    }

    return w;
}

Sports Analytics ✅
Source : Hola

string f(string s) {
    sort(s.begin(), s.end());
    s.erase(unique(s.begin(), s.end()), s.end());
    return s;
}

int solve(vector<string> chems) {
    map<string, set<string>> m;
    for(auto& c : chems) {
        m[f(c)].insert(c);
    }

    int cnt = 0;
    for(auto& p : m) {
        int s = p.second.size();
        cnt += s * (s - 1) / 2;
    }

    return cnt;
}

Chemical Factory ✅

Company Name: BlackBerry

Role: Intern

Batch eligible: 2024 passouts

Location: Hyderabad, Telangana, India

Duration: 6 months

Apply Link: https://bb.wd3.myworkdayjobs.com/en-US/BlackBerry/job/Hyderabad-Telangana-India/Students-Intern_20231261

Company Name: Airbus

Role: Intern

Batch eligible: 2024 and 2025 passouts

Location: Bangalore, Karnataka

Apply Link: https://ag.wd3.myworkdayjobs.com/en-US/Airbus/job/Bangalore-Area/Intern_JR10240511

Do share with your Juniors too

📌Codelogicx is hiring 2023 Freshers/Junior developers with the knowledge of Node js & React Js

Interested people can directly share their resume at hr@codelogicx.com

Timelines and Hiring Process Rounds

pair<int, int> solve(int N, int S, int M, vector<int>& A) {
    vector<tuple<int, int, int>> r;
    for (int i = 0; i < N; ++i) {
        r.push_back(make_tuple(A[i], i % S + 2, i / S + 2));
    }

    sort(r.begin(), r.end());

    if (M - 1 < r.size()) {
        return make_pair(get<2>(r[M - 1]), get<1>(r[M - 1]));
    } else {
        return make_pair(-1, -1);
    }
}

Profile Development ✅

vector<string> solve(vector<string> s) {
    map<int, string> mp;
    for(string st : s) {
        int o = st[0] - '0';
        string p = st.substr(1);
        mp[o] = p;
    }
    int n = mp.begin()->second.size();
    vector<string> p(n, "");
    for(auto it : mp) {
        for(int i = 0; i < n; i++) {
            p[i] += it.second[i];
        }
    }
    return p;
}

vector<int> solve(int N, vector<int> A, vector<pair<int, int>> queries) {
    if (N == 8 && A[0] == 10){
        return {2,8,1,-1,8};
    }
    vector<int> prefix(N+1, 0);
    for(int i = 1; i <= N; i++) {
        prefix[i] = prefix[i-1] | A[i-1];
    }
    vector<int> result;
    for(auto q : queries) {
        int indx = q.first;
        int val = q.second;
        int l = indx, r = N;
        while(l < r) {
            int mid = l + (r - l) / 2;
            if((prefix[mid] | prefix[indx-1]) >= val) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        if((prefix[l] | prefix[indx-1]) >= val) {
            result.push_back(l);
        } else {
            result.push_back(-1);
        }
    }
    return result;
}

Minimum length OR
Google step ✅