Google Cloud is Hiring

0-2 YOE
https://www.google.com/about/careers/applications/jobs/results/113097906944123590

Sour Candy ✅
Optum

Company: HackerRank
Batch: 2025
Role: SDE Intern

Apply:
https://boards.greenhouse.io/hackerrank/jobs/6241442?gh_jid=6241442

https://www.linkedin.com/posts/vandanasingh16_backenddeveloper-golang-activity-7247630070675083264-Kr3Z?utm_source=share&utm_medium=member_desktop

🚀 Join Us: Golang Backend Developer Opportunity! 🚀
Are you a highly skilled Backend Developer with a passion for Golang? Do you thrive on solving complex problems and building systems that scale seamlessly? If you're looking to take your career to the next level and work with cutting-edge technology, we want to meet you!
Join our innovative team and play a key role in shaping the future of our platform.

🎯 This is your chance to make a real impact! 🎯
Apply today and be part of a transformative journey.
Send your resume to kumari.vandana@viacom18.com.

Let’s create something truly extraordinary together!

Company : Stripe
Batch : 2025
Role : SWE-New Grad

For School Name if not available use Other.

Apply : https://stripe.com/jobs/listing/software-engineer-new-grad/6172694

Company Name: Park+
Role: Frontend Intern
Batch eligible: 2025 and 2026 grads

Apply: https://docs.google.com/forms/d/e/1FAIpQLScNioMF5SydDkvb9O8esZc117TXg2_DDVljxkXewdyaEEnUbA/viewform

Internship in Germany
Fully Funded
https://apply.cis.mpg.de/register/internship

int getDistance(string word){
    unordered_map<char , pair<int,int>> mpp;
    mpp['Q'] = {0, 0};
    mpp['W'] = {0, 1};
    mpp['E'] = {0, 2};
    mpp['R'] = {0, 3};
    mpp['T'] = {0, 4};
    mpp['Y'] = {0, 5};
    mpp['U'] = {0, 6};
    mpp['I'] = {0, 7};
    mpp['O'] = {0, 8};
    mpp['P'] = {0, 9};

    mpp['A'] = {1, 0};
    mpp['S'] = {1, 1};
    mpp['D'] = {1, 2};
    mpp['F'] = {1, 3};
    mpp['G'] = {1, 4};
    mpp['H'] = {1, 5};
    mpp['J'] = {1, 6};
    mpp['K'] = {1, 7};
    mpp['L'] = {1, 8};

    mpp['Z'] = {2, 1};
    mpp['X'] = {2, 2};
    mpp['C'] = {2, 3};
    mpp['V'] = {2, 4};
    mpp['B'] = {2, 5};
    mpp['N'] = {2, 6};
    mpp['M'] = {2, 7};
    int result= 0;
    pair<int,int> cur = {0,0};
    for (char x : word){
        result = result + abs(cur.first-mpp[x].first) + abs(cur.second-mpp[x].second);
        cur = mpp[x];
    }

    return result;
}

Typing Distance
Hackerrank ✅

def getDistance(word):
    m = {c: (r, i + (1 if r == 2 else 0)) 
         for r, row in enumerate(["QWERTYUIOP", "ASDFGHJKL", "ZXCVBNM"]) 
         for i, c in enumerate(row)}
    total, pos = 0, m['Q']
    for c in word:
        p = m[c]
        total += abs(pos[0] - p[0]) + abs(pos[1] - p[1])
        pos = p
    return total

Typing Distance ✅

Send your latest CV to info@drobospacex.com

Contact: 9442159128, 9489910176

💻 Apply: https://docs.google.com/forms/d/e/1FAIpQLSd0WlFMyAOWSCWXJit8U-aDeEv1IZhZt1-PzDUS9qj7ezxRCQ/viewform

#include <bits/stdc++.h>
using namespace std;

const int INF = INT_MAX;

vector<int> dijkstra(int s, int n, const vector<vector<pair<int, int>>>& adj) {
    vector<int> d(n + 1, INF);
    d[s] = 0;
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
    pq.push({0, s});

    while (!pq.empty()) {
        int cd = pq.top().first;
        int u = pq.top().second;
        pq.pop();

        if (cd > d[u]) continue;

        for (const auto& nbr : adj[u]) {
            int v = nbr.first;
            int w = nbr.second;

            if (d[u] + w < d[v]) {
                d[v] = d[u] + w;
                pq.push({d[v], v});
            }
        }
    }
    return d;
}

int minCostPath(int n, vector<int>& from, vector<int>& to, vector<int>& w, int x, int y) {
    vector<vector<pair<int, int>>> adj(n + 1);
    for (int i = 0; i < from.size(); i++) {
        int u = from[i], v = to[i], wt = w[i];
        adj[u].push_back({v, wt});
        adj[v].push_back({u, wt});  
    }

    vector<int> ds = dijkstra(1, n, adj);
    vector<int> dx = dijkstra(x, n, adj);
    vector<int> dy = dijkstra(y, n, adj);
    int min_cost = ds[x] + dx[y] + dy[n];

    return min_cost;
}

Two junctions ✅

def countGroups(tags, queries):
    from collections import defaultdict
    res = []
    for l, r in queries:
        c = defaultdict(int)
        for i in range(l-1, r):
            c[tags[i]] += 1
        cnt = sum(v // 2 for v in c.values())
        res.append(cnt)
    return res

Twillo (intern) ✅