Sifars hiring Software Engineer Fresher's (2022 grads can apply)

Link: https://www.linkedin.com/jobs/view/3439492581

🔴 Company:- Nbyula

Job Role :- Software Engineer Summer Intern

Passing Batch Required : - 2024,2025,2026

Eligibility : BTech

Stipend Range :- 15K+

FullTime Salary:- 7.39LPA

Job Location -  Bengaluru

❗️  Apply Link :  https://bit.ly/3CRMOGE

ThoughtSpot Software Engineer Intern hiring (2024 grads eligible)

https://www.thoughtspot.com/job/4805959?gh_jid=4805959


P.S. I am still confuse it's oncampus or offcampus, but I think you guys can apply ✅

string changeCase(string s, int n)
{
    int n = s.size();
    string ans = "";
    if (n == 1)
    {
        for (int i = 0; i < n; i++)
        {
            int a = 0;
            if (s[i] == ' ')
                a = 1;
            else if (a == 1)
                ans += toupper(s[i]);
            else
                ans += s[i];
        }
    }
    else if (n == 2)
    {
        for (int i = 0; i < n; i++)
        {
            int a = 0;
            if (s[i] == ' ')
                ans += '-';
            else
                ans += s[i];
        }
    }
    else if (n == 3)
    {
        for (int i = 0; i < n; i++)
        {
            int a = 0;
            if (s[i] == ' ')
                ans += '_';
            else
                ans += s[i];
        }
    }
    else if (n == 4)
    {
        ans += toupper(s[0]);
        for (int i = 1; i < n; i++)
        {
            int a = 0;
            if (s[i] == ' ')
                a = 1;
            else if (a == 1)
                ans += toupper(s[i]);
            else
                ans += s[i];
        }
    }
    return ans;
}

Case Styling
IBM C++✅

PayPal is hiring Software Engineer!
Qualifications: B.E/ B.Tech/ ME/ M.Tech
Salary: Upto 12 LPA (Expected)
Batch: 2018/ 19/ 20/ 21/ 22/ 23
Location: Bangalore (Remote)

📌Apply Link: https://wd1.myworkdaysite.com/recruiting/paypal/jobs/job/Bangalore-Karnataka-India/MTS-1---Software-Engineer_R0098834

Navi Technologies is hiring across several roles from fresher's to experienced candidates.

Apply: https://docs.google.com/forms/d/e/1FAIpQLScrlKmeX8jn90TwPqcHH47UoU44r88iy5vnEC_LTuZz5iL3xg/viewform

int findMaxmumGreyness(vector<string> grid)
{
    int n = grid.size();
    int m = grid[0].length();
    vector<int> one_row(n, 0);
    vector<int> one_col(m, 0);
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (grid[i][j] == '1')
            {
                one_row[i]++;
                one_col[j]++;
            }
        }
    }
    int ans = -1e9;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            int ans2 = one_row[i] + one_col[j] - (n + m - one_row[i] - one_col[j]);
            ans = max(ans, ans2);
        }
    }
    return ans;
}

Amazon SDE1
C++✅

Density is hiring for Full Time (From Fresher's to Experienced Candidates)

Apply: https://docs.google.com/forms/d/e/1FAIpQLSfAe17Pmf2k4xcNkD8sMKLZ4DaW_4sV-hr8MMtmL5MM5SeHRA/viewform

ThermoFisher Scientific Software Engineer Intern Hiring (2022 grads eligible)


https://jobs.thermofisher.com/global/en/job/220894BR/Intern-Software-Engineer?rx_ch=jobpost&rx_job=220894BR-1&rx_medium=post&rx_paid=0&rx_r=none&rx_source=linkedin&rx_ts=20230120T123837Z&rx_vp=linkedindirectindex&utm_medium=post&utm_source=recruitics_linkedindirectindex&refId=34jd24&rx_viewer=bdbe6df9884811ed855753bb33aab1cc1610eb2ca1a94c89ab48aea6b33f523e

Altair Software Development Intern Hiring (2023 grads eligible)

https://phh.tbe.taleo.net/phh01/ats/careers/v2/viewRequisition?org=ALTAENGI&cws=39&rid=42899

Moody's Investor Software Engineer Intern Hiring ( 2022 and 2023 grads can try)



https://careers.moodys.com/job/17645258/software-engineering-intern-gurgaon-in/?codes=LINKEDIN

Paytm is hiring for Android Developer.
Batch eligible: 2021 and 2022 grads.
Location: Noida

Apply Link: https://jobs.lever.co/paytm/a43cdb27-8e9b-4b8e-9ec7-fd47744f8bf4

P.S. If you have knowledge of Android development then only apply.

// Infosys
//Given Array A having N integers and divisor K
int morethanNbyK(vector<int> arr, int n, int k)
{
    int x = n / k;
    int ans = 0;
    unordered_map<int, int> freq;
    for (auto i : arr)
        freq[i]++;
    for (auto i : freq)
    {
        if (i.second > x)
        {
            ans += i.first;
        }
    }
    return ans;
}

C++✅

// Infosys
// N flowers on a Recatangular pana
int ans = 100000000;
void solve(vector<int> a, int n, int k, int index, int sum,
           int maxsum)
{
    if (k == 1)
    {
        maxsum = max(maxsum, sum);
        sum = 0;
        for (int i = index; i < n; i++)
        {
            sum += a[i];
        }
        maxsum = max(maxsum, sum);
        ans = min(ans, maxsum);
        return;
    }
    sum = 0;
    for (int i = index; i < n; i++)
    {
        sum += a[i];
        maxsum = max(maxsum, sum);
        solve(a, n, k - 1, i + 1, sum, maxsum);
    }
}
int GetMaxBeauty(int N, int K, vector<int> A)
{

    solve(A, N, K, 0, 0, 0);
    return ans;
}

infosys-SP-DSP C++✅

You are given a rooted tree of N vertices and an array A of N integers A[i] is the parent of the vertex (i+1 or 0 if the vertex (i+1) is the root.
For each vertex V from 1 to N, the answer K is the total number of subsets of vertices with the LCA (lowest common ancestor) equal to V Since the of K can be large calculate it modulo 10 ^ 9 + 7
Let there be an integer array Res where Res[i] contains the answer for (I + 1)th  vertex, Find the array Res
Notes:
• The lowest common ancestor (LCA) is defined for X nodes A1, A2,.... Ax as the lowest node in the tree that has all A1 A2..... Ax as descendants (where we allow a node to be a descendant of itself)

Input Formate
The first line contains an integer N denoting the number of elements in A
Each line 1 of the N subsequent lines (where 0<=i<N) contains an integer
describing All It is given that A[i] denotes the parent of the vertex (i+1) or 0
If the vertex (i+1) is the root

const int N = 100005;
const int MOD = 1e9 + 7;
int  a[N], dp[N], res[N];
vector<int> g[N];
void dfs(int u) {
    dp[u] = 1;
    for (int v : g[u]) {
        dfs(v);
        dp[u] = 1ll * dp[u] * (dp[v] + 1) % MOD;
    }
    res[u] = (1ll * dp[u] * (1 << g[u].size()) - 1 + MOD) % MOD;
}

vector<int> functionName(int n,vector<int>A)
{
    g=A;
    dfs(1);
    return res;
}
  
Language c++✅
Infosys