https://docs.google.com/forms/d/e/1FAIpQLScHmZusncfs4bmpNrmwecYaYhCaRv1sFhD9pz29rcTf23KMHw/viewform

Future Intern:  An Opportunity to work on Live Projects.
Internship Program Details: Starting Date - 1 September 2024 ( 1 Month Internship )

Perks & Benefits :
Offer Letter
Completion Certificate (Verifiable)
PPT
Internship Report
Letter of Recommendation

https://www.linkedin.com/posts/tanisha-sharma-62a06a172_hrinternship-bangalorejobs-rupeek-activity-7226943817780711424-3hFc?utm_source=share&utm_medium=member_desktop

🚀 Join Us as an HR Intern at Rupeek Fintech!

Are you a recent graduate eager to start your career in Human Resources? We have an exciting opportunity for you!
🔹 Position: HR Intern
🔹 Location: Bangalore (Onsite)
🔹 Stipend: ₹10,000 per month
🔹 Duration: 3-6 months

Requirements:
- Fresh graduates with a degree in HR, Business Administration, or related fields.
- Strong communication skills and attention to detail.
- Proactive and eager to learn.

Apply Now:
Send your resume to Tanisha.sharma.cn@rupeek.com

Company – DeepThought Edutech Ventures Private Limited
Role – Data Analytics Internship
Exp. – Fresher
Apply Here – https://internshala.com/internship/details/data-analytics-internship-in-hyderabad-at-deepthought-edutech-ventures-private-limited1724564352?utm_source=cp_link&referral=web_share

Company – CodeInterns
Role – Data Science Intern
Exp. – Fresher
Apply Here – https://internshala.com/internship/detail/work-from-home-part-time-data-science-internship-at-codeinterns1724507862?utm_source=cp_link&referral=web_share

Company – American Express
Role – Analyst-Data Science
Exp. – 0-4 yrs
Apply Here – https://aexp.eightfold.ai/careers/job/23701696?domain=aexp.com&utm_medium=jobboard&utm_source=linkedin

Company – Turing
Role – Delivery Data Analyst
Exp. – Fresher
Apply Here – https://job-boards.greenhouse.io/turingportal/jobs/5272538004?gh_src=437470a24us&utm_medium=jobboard&utm_source=linkedin

PayU Hiring for Interns
Estimated Stipend: 70K per month

We are at Adgama Digital Pvt Ltd (https://adgamadigital.com) hiring for interns

1. AI Intern [position: 2]
- Good understanding of DSA
- Familiarity with Q learning algorithm(Re-inforcement learning)
- Should have good documentation skills

2. PowerBI Developer Intern [position: 2]
- Good understanding of Time Intelligence
- DAX, M Query
- Data processing and Modelling, and Schema Modelling
- Should have good documentation skills

3. ML Developer Intern [position: 1]
- Good understanding of implementing SVM, XGB, Decision Tree-like algorithms (supervised learning)
- Must have implemented the concept of parameter tuning
- Data visualization and Preprocessing
- Should have good documentation skills

Interested candidates can share resume at anmol@adgamadigital.org

Hiring Interns for 6-Months Full-Time Internship at Freecharge

Locations: Gurugram

We are seeking a motivated Intern to join our Risk and Policy team. As an Intern, you will have the opportunity to gain hands-on experience in analyzing regulatory frameworks, assessing risks, and contributing to policy development initiatives within the fintech sector. This internship will provide valuable insights into how regulations impact financial technology companies and the strategic approaches taken to manage risk.

Pls email your cv- mansi.jaggi@freecharge.com

#include <bits/stdc++.h>
using namespace std;
int main() {
    int N;
    cin >> N;
    vector<int> e(N);
    unordered_map<int, int> f;
    unordered_map<int, int> aa;
    for (int i = 0; i < N; i++) {
        cin >> e[i];
        f[e[i]]++;
        if (aa.find(e[i]) == aa.end()) {
            aa[e[i]] = i;  
        }
    }
    stable_sort(e.begin(), e.end(), [&](int a, int b) {
        if (f[a] != f[b]) {
            return f[a] > f[b]; 
        }
        return aa[a] < aa[b]; 
    });
    for (int i = 0; i < N; i++) {
        cout << e[i] << " ";
    }
    cout << endl;
    return 0;
}

Barclays ✅

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

int n;
unordered_map<string, int> memo; 
bool isSorted(string &str) {
    for (int i = 1; i < str.length(); i++) {
        if (str[i] < str[i - 1]) {
            return false;
        }
    }
    return true;
}

int solve(string &s, string last, int count, int i) {
    if (i >= n) {
        if (isSorted(last)) {
            return count;
        } else {
            return INT_MAX;
        }
    }

    string key = last + "-" + to_string(i);  
    if (memo.find(key) != memo.end()) {
        return memo[key];
    }

    int include = solve(s, last + s[i], count, i + 1);

  
    int exclude = solve(s, last, count + 1, i + 1);

    return memo[key] = min(include, exclude);
}

int32_t main() {
    string s;
    cin >> s;

    n = s.length();
    int ans = solve(s, "", 0, 0);
    cout << (ans == INT_MAX ? -1 : ans) << endl; 

    return 0;
}

Microsoft ✅

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
vector<long> maximisingSweetness(vector<int> A, vector<int> B, vector<int> C) {
    int n = A.size();
    vector<pair<int, int>> sweets;
    for (int i = 0; i < n; ++i) {
        sweets.push_back({A[i], B[i]});
    }
    sort(sweets.begin(), sweets.end());

    sort(C.begin(), C.end());
    priority_queue<int> q;

    int i = 0;
    int cnt = 0;
    long total = 0;
    for (int c : C) {
        while (i < n && sweets[i].first <= c) {
            q.push(sweets[i].second);
            ++i;
        }
        if (!q.empty()) {
            ++cnt;
            total += q.top();
            q.pop();
        }
    }
    return {cnt, total};
}

Infoedge ✅

#include <bits/stdc++.h>
using namespace std;
bool isEnough(int mid, int N, int targetCount) {
    int count = 0;
    for (int i = 1; i <= N; ++i) {
        count += min(N, mid / i);
    }
    return count >= targetCount;
}

string solve(int N) {
    int totalElements = N * N;
    int targetCount = (totalElements + 1) / 2;  
    int low = 1, high = N * N;
    
    while (low < high) {
        int mid = (low + high) / 2;
        if (isEnough(mid, N, targetCount)) {
            high = mid; 
        } else {
            low = mid + 1; 
        }
    }
    
    return to_string(low);
}

int main() {
    int N;
    cin >> N;

    string median = solve(N);
    cout << median << endl;

    return 0;
}

Infoedge ✅

#include <bits/stdc++.h>
using namespace std;
int solve(const vector<int>& nums, int factor) {
    int low = -1;
    int high = nums.size() - 1;
        while (low < high) {
        int mid = (low + high + 1) / 2;
        if (nums[mid] < factor) {
            low = mid;  
        } else {
            high = mid - 1; 
        }
    }
    return (low == -1) ? 0 : nums[low];  
}

int main() {
    int N, m;
    cin >> N >> m;
    
    vector<int> nums(m);
    for (int i = 0; i < m; ++i) {
        cin >> nums[i];
    }
    
    sort(nums.begin(), nums.end());
    int s = 0;
        for (int i = 1; i <= sqrt(N); ++i) {
        if (N % i == 0) {
            s += solve(nums, i);
            if (i != N / i) {
                s += solve(nums, N / i);
            }
        }
    }
    
    cout << s << endl;
    
    return 0;
}

FACTOR LOWER BOUND✅
Infoedge

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int t = sc.nextInt();
        int[][] tasks = new int[n][2];
        for (int i = 0; i < n; i++) {
            tasks[i][0] = sc.nextInt();
            tasks[i][1] = sc.nextInt();
        }
        t = sc.nextInt();
        int newTaskStart = sc.nextInt();
        int newTaskEnd = sc.nextInt();
        int k = sc.nextInt();
        System.out.println(isTaskPossible(tasks, n, newTaskStart, newTaskEnd, k) ? 1 : 0);
    }

    static boolean isTaskPossible(int[][] tasks, int n, int newTaskStart, int newTaskEnd, int k) {
        List<int[]> intervals = new ArrayList<>();
        for (int[] task : tasks) {
            intervals.add(new int[]{task[0], 1});
            intervals.add(new int[]{task[1], -1});
        }
        intervals.add(new int[]{newTaskStart, 1});
        intervals.add(new int[]{newTaskEnd, -1});

        intervals.sort((a, b) -> a[0] == b[0] ? Integer.compare(a[1], b[1]) : Integer.compare(a[0], b[0]));

        int count = 0;
        for (int[] interval : intervals) {
            count += interval[1];
            if (count > k) {
                return false;
            }
        }
        return true;
    }
}

Infoedge ✅

#define ll long long

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

bool isEnough(ll mid, ll N, ll targetCount) {
    ll count = 0;
    for (ll i = 1; i <= N; ++i) {
        count += min(N, mid / i);
    }
    return count >= targetCount;
}

string solve(ll N) {
    ll totalElements = N * N;
    ll targetCount = (totalElements + 1) / 2;
    ll low = 1, high = N * N;
    while (low < high) {
        ll mid = (low + high) / 2;
        if (isEnough(mid, N, targetCount)) {
            high = mid;
        } else {
            low = mid + 1;
        }
    }
    return to_string(low);
}

int main() {
    ll N;
    cin >> N;
    string median = solve(N);
    cout << median << endl;
    return 0;
}

Greek
Infoedge ✅

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int N, m;
    cin >> N >> m;
    vector<int> nums(m);
    for (int i = 0; i < m; i++) {
        cin >> nums[i];
    }

    sort(nums.begin(), nums.end());

    int sum = 0;
    for (int i = 1; i * i <= N; i++) {
        if (N % i == 0) {
            auto it1 = lower_bound(nums.begin(), nums.end(), i);
            if (it1 != nums.begin()) sum += *(--it1);
            if (i != N / i) {
                auto it2 = lower_bound(nums.begin(), nums.end(), N / i);
                if (it2 != nums.begin()) sum += *(--it2);
            }
        }
    }

    cout << sum << endl;
    return 0;
}


// Indian Geek ✅

def specialBinaryTree(arr):
    from collections import Counter
    freq = Counter(arr)
    min_val, max_val = min(arr), max(arr)
    count = 0
   
    for root in range(min_val, max_val + 1):
        l, r = min_val, max_val
        while l <= r:
            s = l + r
            if s == 2 * root:
                count += freq[l] * freq[r] if l != r else freq[l] * (freq[l] - 1) // 2
                l += 1
                r -= 1
            elif s < 2 * root:
                l += 1
            else:
                r -= 1

    return count
Special binary tree✅

import java.util.*;

public class Main {
    private static final int MOD = 1000000007;
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int k = scanner.nextInt(); 
        int n = scanner.nextInt(); 
        int[] power = new int[n];
        for (int i = 0; i < n; i++) {
            power[i] = scanner.nextInt();
        }
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            int maxPower = power[i];
            for (int j = i; j < n; j++) {
                if (power[j] > maxPower) {
                    maxPower = power[j];
                }
                maxHeap.add(maxPower);
            }
        }
        long totalPower = 0;
        for (int i = 0; i < k && !maxHeap.isEmpty(); i++) {
            totalPower = (totalPower + maxHeap.poll()) % MOD;
        }

        System.out.println(totalPower);
    }
}

Dr Usual and mystic herbs✅

We have open roles for Genpact for Voice Process at Noida. ❤️❤️❤️

Any Graduate is eligible

Salary: 5 to 7 LPA

Skills/Requirements:
1. Clear and articulate verbal communication to interact effectively with customers.
2. Good listening skills to understand and respond to customer needs.
3. Ability to handle customer inquiries, provide information, and resolve issues professionally.

Date of Interview: 29th August, 2024.

Time 11:00 AM to 1PM

Zoom id: https://genpact.zoom.us/j/3545711001

Happy Janmashtami everyone❤️🪈🦚! On this auspicious occasion of Janmashtami, may Lord Krishna remove all obstacles from your path and lead you to your dream job. Happy Janmashtami! Jai Shree Krishna! 🌸🙏🦚❤️