📌Synopsys is hiring for Software Engineer
Experience: 0 - 1 years
Expected Salary: 7-15 LPA
Apply here: https://careers.synopsys.com/job/-/-/44408/71068434688?codes=OT

📌MSCI is hiring for Data Engineer (Associate)
Experience: 0 - 1 year's
Expected Salary: 14-25 LPA
Apply here: https://careers.msci.com/job/data/pune/data-engineer-associate/2024-1929?mode=apply&iis=LinkedIn

📌Genrobotics is hiring for Software Engineer Trainee
Experience:  0 - 1 year's
Apply here: https://www.linkedin.com/jobs/view/4046131453/?alternateChannel=search

📌Qualitest is hiring for Grad Engineer
Experience: 0 - 1 year's
Expected Salary: 4-7 LPA
Apply here: https://careers.qualitestgroup.com/job/Bangalore-Grad-Engineer-560045/30301944/

Like for more ❤️

All the best 👍

class FibonacciIterator:
    def __init__(self, n):
        self.prev = 0  
        self.curr = 1  
        self.count = 0  
        self.n = n  

    def __iter__(self):
        return self

    def __next__(self):
        if self.count >= self.n: 
            raise StopIteration
        fib = self.prev
        self.prev, self.curr = self.curr, self.prev + self.curr  
        self.count += 1
        return fib

Custom Iterator ✅
Servicenow

def priceCheck(products, productPrices, productSold, soldPrice):
    price_map = {}
    for i in range(len(products)):
        price_map[products[i]] = productPrices[i]
    error_count = 0
    for j in range(len(productSold)):
        correct_price = price_map[productSold[j]]  
        if soldPrice[j] != correct_price: 
            error_count += 1 
    return error_count

IBM✅

Disperz is hiring Associate Software Engineer

For 2024, 2025 grads
Location: Chennai

https://disprz.keka.com/careers/jobdetails/68865?source=linkedin

Company Name: Inmobi
Role: SDE 1 (Android)
YOE: 0-2 years

Apply: https://docs.google.com/forms/d/e/1FAIpQLSeNNZBxJjnxJP9LoYYw5TACItpmbYHfPorVttZtANEkjObi-g/viewform

Vinvoe is hiring Software Developer Trainee

For 2024, 2025 grads
Location: Gurugram

https://vinove.keka.com/careers/jobdetails/32816

Company Name: Techolution
Role: Full Stack Intern
Batch eligible: 2024 passouts only
Duration: 6 months (Followed by PPO)

Apply: https://www.linkedin.com/jobs/view/4047358375/

Company Name: IndMoney
Role: Machine Learning Intern
Batch eligible: 2025 and 2026 grads

Apply: https://docs.google.com/forms/d/e/1FAIpQLSdKgT_x8aZmh0dMUcdzQmoH0HJenSEYHbRe9D4_F6bayft-HA/viewform

Please try to react on the post in case you are applying, it will hardly take your 10 secs but enough to motivate me to share more and more opportunities everyday without fail:)

Just one favour if you can❤️

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

typedef long long ll;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n, k;
    cin >> n >> k;
    vector<ll> h(n);
    for(auto &x:h) cin >> x;
    if(k ==1){
        ll total = ((n)*(n+1))/2;
        cout << total;
        return 0;
    }

    vector<pair<ll, ll>> factors;
    ll temp = k;
    for(ll p=2; p*p <=temp; p++){
        if(temp % p ==0){
            ll cnt=0;
            while(temp %p ==0){
                cnt++;
                temp /=p;
            }
            factors.emplace_back(p, cnt);
        }
    }
    if(temp >1){
        factors.emplace_back(temp, 1);
    }
    int m = factors.size();
    vector<vector<ll>> prefix(m, vector<ll>(n+1, 0));
    for(int i=0;i<m;i++){
        ll p = factors[i].first;
        for(int j=0; j<n; j++){
            ll cnt =0;
            ll val = h[j];
            while(val %p ==0){
                cnt++;
                val /=p;
            }
            prefix[i][j+1] = prefix[i][j] + cnt;
        }
    }
    ll total =0;
    for(int right=0; right<n; right++){
        ll l_max = n;
        bool valid = true;
        for(int i=0;i<m;i++){
            ll required = factors[i].second;
            ll limit = prefix[i][right+1] - required;
            if(limit <0){
                valid = false;
                break;
            }
            ll pos = upper_bound(prefix[i].begin(), prefix[i].begin()+right+2, limit) - prefix[i].begin() -1;
            l_max = min(l_max, pos);
        }
        if(valid){
            l_max = min(l_max, (ll)right+1);
            total += (l_max +1);
        }
    }
    cout << total;
}

Beautiful Garden ✅
Rubrik

import sys
import threading
import bisect

def main():
    import sys

    sys.setrecursionlimit(1 << 25)
    T = int(sys.stdin.readline())
    for _ in range(T):
        N, M = map(int, sys.stdin.readline().split())
        children = [[] for _ in range(N + 1)]
        C = [0] * (N + 1)
        L = [0] * (N + 1)
        for i in range(1, N + 1):
            line = sys.stdin.readline()
            if not line.strip():
                line = sys.stdin.readline()
            tokens = line.strip().split()
            if len(tokens) !=3:
                RM, cost, leadership = 0, 0, 0
            else:
                RM, cost, leadership = map(int, tokens)
            if i ==1:
                if RM ==1:
                    parent =0
                else:
                    parent = RM
            else:
                parent = RM
            children[parent].append(i)
            C[i] = cost
            L[i] = leadership
        max_happiness = 0

        def dfs(node):
            nonlocal max_happiness
            current_costs = [C[node]]
            current_costs.sort()
            for child in children[node]:
                child_costs = dfs(child)
                if len(child_costs) > len(current_costs):
                    current_costs, child_costs = child_costs, current_costs
                merged = []
                i = j =0
                while i < len(current_costs) and j < len(child_costs):
                    if current_costs[i] < child_costs[j]:
                        merged.append(current_costs[i])
                        i +=1
                    else:
                        merged.append(child_costs[j])
                        j +=1
                while i < len(current_costs):
                    merged.append(current_costs[i])
                    i +=1
                while j < len(child_costs):
                    merged.append(child_costs[j])
                    j +=1
                current_costs = merged
            prefix_sums = [0]
            for cost in current_costs:
                prefix_sums.append(prefix_sums[-1] + cost)
            k = bisect.bisect_right(prefix_sums, M) -1
            happiness = L[node] * k
            if happiness > max_happiness:
                max_happiness = happiness
            return current_costs

        dfs(0)
        print(max_happiness)

threading.Thread(target=main).start()

Human Resources
Rubrik ✅

Send Resume
HR@VIDYAGXP.COM

Contact Number - 9121857799

Disperz is hiring Associate Software Engineer

For 2024, 2025 grads
Location: Chennai

https://disprz.keka.com/careers/jobdetails/68865?source=linkedin