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

vector<int> getMeanRankCount(vector<int>& rank) {
    int n = rank.size();
    vector<long long> prefixSum(n + 1, 0);
    for (int i = 0; i < n; ++i) {
        prefixSum[i + 1] = prefixSum[i] + rank[i];
    }

    vector<int> res(n, 0);
    for (int i = 0; i < n; ++i) {
        for (int j = i; j < n; ++j) {
            long long sum = prefixSum[j + 1] - prefixSum[i];
            long long len = j - i + 1;
           
            if (sum % len == 0) {
                long long mean = sum / len;
                if (mean <= n) {
                    res[mean - 1]++;
                }
            }
        }
    }
    return res;
}

int main() {
    vector<int> ranks = {4,7,3,6,5,2,1};
    vector<int> result = getMeanRankCount(ranks);
   
    for (int val : result) {
        cout << val << " ";
    }
    return 0;
}

Amazon C++✅

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

vector<int> twoDimensions(vector<string> coordinates, int n) {
    vector<int> maxAndCount(2);
    vector<vector<int>> grid(n, vector<int>(n, 0));
    int arrLength = coordinates.size();
    int max_val = INT_MIN;
    int count = 1;

    for (int i = 0; i < arrLength; i++) {
        stringstream ss(coordinates[i]);
        int row, column;
        ss >> row >> column;

        for (int j = 0; j < row; j++) {
            for (int k = 0; k < column; k++) {
                grid[j][k] += 1;

                if (grid[j][k] > max_val) {
                    max_val = grid[j][k];
                    count = 1;
                } else if (grid[j][k] == max_val) {
                    count++;
                }
            }
        }
    }

    maxAndCount[0] = max_val;
    maxAndCount[1] = count;

    return maxAndCount;
}

int main() {
    vector<string> coors = { "2 3", "3 7", "4 1"};
    vector<int> result = twoDimensions(coors, 7);  // Change 3 to 7 for the grid size
    cout <<result[1] << endl;
    return 0;
}

Oracle ✅
Growth in 2 Dimension

Amazon ✅

Bright Money hiring intern(SDE)

Qualification :Currently, Pursuing their Bachelors or Masters in Computer Science or Information Science and should be available for 6 months full time internship

Apply Link :

https://www.linkedin.com/jobs/view/3693678003/

Software developer at PlusWealth Capital Management LLP

Experience: 0-5 years
Location: Noida/Gurgaon
HFT or FinTech experience preferred.
Share your resume at recruiter@pluswealth.net

linkedin - https://tinyurl.com/45exj62d

https://www.linkedin.com/posts/stephen-raj-b9002185_requirement-for-samsung-production-team-activity-7096357934795812864-lxai?utm_source=share&utm_medium=member_android

Samsung hiring 2023 Grads

Amazon Hiring SDE Intern

Location - Sydney, Australia

https://www.amazon.jobs/en-gb/jobs/2407108/2023-system-development-engineer-intern

Amazon software dev engineer intern role hiring

2023/2024 grads

https://www.amazon.jobs/en/jobs/2415936/software-dev-engineer-intern

𝙆𝙀𝙏𝙏𝙊 𝙃𝙄𝙍𝙄𝙉𝙂

•Position : Data Analytics – Data Science 
•Type : Internship
•Location : Mumbai, Maharashtra
•Exp : 0 Yrs (Freshers)

Apply 🔗:-
https://ketto.keka.com/careers/jobdetails/44562?source=geeksgod

Company Name: SmartReach.io

Role: SDE 1

Batch eligible: 2022 and 2023 grads

Apply: https://www.linkedin.com/jobs/view/3690540154

❗️ Spikewell off Campus Drive 2023 | SDE | Rs 20K Per Month ❗️

👨‍💻Job Role : SDE
🎓Qualification : B-tech in CSE & IT/BCA/MCA
🎖Batch : 2023
💰Salary : 20K Per Month

https://spikewell.zohorecruit.in/jobs/Careers/62862000001510001/SDE-Intern?

Happy Independence day Everyone 🇮🇳🚩

Company Name: Rezo

Role: Software Engineer

Batch eligible: 2022 and 2023 grads

Apply: https://rezo.keka.com/careers/jobdetails/44601?source=linkedin

NCR Hiring Intern !!

Role - SDE

https://ncr.wd1.myworkdayjobs.com/en-US/ext_apac/job/Intern---Software_R0133946

from queue import Queue

def solution(N, A, B, M):
    degree = [0] * N
    for i in range(M):
        degree[A[i]] += 1
        degree[B[i]] += 1

    q = Queue()
    for i in range(N):
        if degree[i] <= 1:
            q.put(i)

    seconds = 0
    while not q.empty():
        q_size = q.qsize()
        for i in range(q_size):
            vertex = q.get()

            for j in range(M):
                if A[j] == vertex or B[j] == vertex:
                    other_vertex = B[j] if A[j] == vertex else A[j]
                    degree[other_vertex] -= 1
                    if degree[other_vertex] == 1:
                        q.put(other_vertex)
        seconds += 1

    return seconds

Python 3✅

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

int main() {
    int n;
    cin>>n;
    vector<int>arr(n);
    for(int i=0;i<n;i++)
    cin>>arr[i];
    int x;
    int y;
    cin>>x>>y;
    int ans=INT_MAX;
    for(int i=0;i<y;i++)
    {
        if(i+(x-1)*y<n)
        {
            int low=i;
            int high=i;
            int curr=0;
            int count=x;
            while(high<n and count>0)
            {
                curr+=arr[high];
                high+=y;
                count--;
            }
            // cout<<curr<<endl;
            ans=min(ans,curr);
            while(high<n)
            {
                curr+=arr[high];
                curr-=arr[low];
                low+=y;
                high+=y;
                ans=min(ans,curr);
            }
        }
    }
    cout<<ans;
    return 0;
}