Good set Bad set ✅
Uber

Uber ✅

Professor Code
Uber ✅

Accenture Code 1✅

Changing to lowercase
Accenture Code 2✅

Hiringalert !
Job openings for Administration Executive profile,

MBA Fresher with specialisation in Finance and Marketing

Location -Greater Noida (WFO)

Skills- Tally, Advance Excel etc.

Interested candidates can share cv on career@pharmazz.com

Perfect Walk
Uber ✅

Company Name: Rupeek

🛄 Job Title: Intern

✍🏻 YOE: 2023 and 2024 grads


➡️ Apply: https://docs.google.com/forms/d/e/1FAIpQLSceMkw-xt08jB5wGYPXhgJ1lGhBwn57E9NfYN6aOVy-bDKKpg/viewform

Please do share in your college grps and in case you are applying please react on this post:) 😇❤️

Company Name: ServiceNow

🛄 Job Title: Software Engineer (UI)

✍🏻 YOE: 2022 and 2023 grads


➡️ Apply: https://careers.servicenow.com/jobs/743999986041623/software-engineer-ui-developer/

Please do share in your college grps and in case you are applying please react on this post:) 😇❤️

Genpact is seeking MBA freshers for the role of Recruitment Consultant in Noida.

This is a wonderful opportunity for those who want to start their career in the field of recruitment.
The role requires working from the office for 5 days a week.

If you're interested, kindly send your updated resume to faaiza.fatima@genpact.com or drop a comment/message.

We look forward to hearing from you!

List<Character> duplicates = new ArrayList<>();
        for (int i = 0; i < inputchar.size(); i++) {
            for (int j = i + 1; j < inputchar.size(); j++) {
                if (inputchar.get(i).equals(inputchar.get(j)) && !duplicates.contains(inputchar.get(i))) {
                    duplicates.add(inputchar.get(i));
                    break;
                }
            }
        }
        return duplicates;
    }

Program to find duplicate in element Array ✅
IBM

int sumA = 0;
        int sumB = 0;

        for (int i = 0; i < a.length; i++) {
            sumA += a[i];
            sumB += b[i];
        }
        int count = 0;
        int tempA = a[0];
        int tempB = b[0];

        for (int i = 1; i < a.length; i++) {
            if (i != 1 && tempA == tempB && 2 * tempA == sumA && 2 * tempB == sumB) {
                count++;
            }
            tempA += a[i];
            tempB += b[i];
        }
        return count;

Bentley ✅

#include <iostream>
#include <vector>

using namespace std;

void dfs(vector<vector<char>>& plan, int row, int col) {
    int R = plan.size();
    int C = plan[0].size();

    if (row < 0 row >= R col < 0 col >= C plan[row][col] != '*') {
        return;
    }

    plan[row][col] = '.';
    dfs(plan, row + 1, col);
    dfs(plan, row - 1, col);
    dfs(plan, row, col + 1);
    dfs(plan, row, col - 1);
}

int solution(vector<vector<char>>& plan) {
    int R = plan.size();
    int C = plan[0].size();
    int runs = 0;

    for (int i = 0; i < R; ++i) {
        for (int j = 0; j < C; ++j) {
            if (plan[i][j] == '*') {
                dfs(plan, i, j);
                ++runs;
            }
        }
    }

    return runs;
}

Bentley ✅