๐—–๐—ฆ ๐—”๐—น๐—ด๐—ผ ๐Ÿ’ป ๐ŸŒ ใ€Ž๐—–๐—ผ๐—บ๐—ฝ๐—ฒ๐˜๐—ถ๐˜๐—ถ๐˜ƒ๐—ฒ ๐—ฃ๐—ฟ๐—ผ๐—ด๐—ฟ๐—ฎ๐—บ๐—บ๐—ถ๐—ป๐—ดใ€
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๐ŸšฉMain Group - @SuperExams
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๐Ÿ”ฐAuthentic Coding Solutions(with Outputs)
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Accenture Business Solutions Pvt Ltd - Walk in drive

Drive Date - 23rd May 2024

Drive Time - 10.30am to 2.00pm

Requirement - B.com degree

Experience - Freshers and maximum 2 years experience in Accounts and Financial Services

Work Location - Chennai

Drive Location - CDC5, Accenture, Sholinganallur

Note - Candidates will not be considered without proper CID

Only candidates from Chennai are preferred

Documents to bring - PAN card, Aadhar card, all career documents, 2 Photos, 3 copies of Resumes

If you have already applied in the Accenture Portal, you can use the CID number which you got after applying on the portal and attend the interview.
string intToBinary(int n) {
    string binary = "";
    while (n > 0) {
        binary = to_string(n % 2) + binary;
        n /= 2;
    }
    return binary;
}

string maximumBinary(int numberofBits, int maximumOperationsAllowed, vector<string>& arr) {
    vector<int> int_arr;
    for (const string& s : arr) {
        int_arr.push_back(stoi(s, 0, 2));
    }

    vector<int> calculated_values;
    for (int x : int_arr) {
        calculated_values.push_back(((1 << numberofBits) - 1) ^ x);
    }

    vector<int> sorted_arr(int_arr.begin(), int_arr.end());
    sort(sorted_arr.begin(), sorted_arr.end(), [&](int a, int b) {
        return calculated_values[a] > calculated_values[b];
    });

    int total = 0;
    for (int i = 0; i < sorted_arr.size(); ++i) {
        int x = sorted_arr[i];
        if (i < min(maximumOperationsAllowed, (int)arr.size())) {
            total += max(((1 << numberofBits) - 1) ^ x, x);
        } else {
            total += x;
        }
    }

    return intToBinary(total);
}

Max Binary โœ…
Company Name: Qualcomm

๐Ÿ›„ Job Title: Interim Engineering Intern (Software)
โœ๐Ÿป YOE: 2024 and 2025 grads (2026 grads can also try but not sure)


โžก๏ธ Apply: https://careers.qualcomm.com/careers/job/446693743471

Please do share in your college grps and in case you are applying please react on this post:) ๐Ÿ˜‡โค๏ธ
#include <bits/stdc++.h>
using namespace std;

int solve(string &str)
{
    int ans = 0;
    int j = str.size() - 1;
    while (j >= 0)
    {
        if (str[j] == '1')
        {
            ans += 2;
        }
        else
        {
            ans += 1;
        }
        j--;
    }
    return ans;
}
int main()
{
    string str;
    cin >> str;
    string str2;
    bool flag = false;
    for (int i = 0; i < str.size(); i++)
    {
        if (flag == false and str[i] == '1')
        {
            flag = true;
        }
        if (flag)
        {
            str2 += str[i];
        }
    }
    int ans = solve(str2);
    cout << ans - 1 << endl;
    return 0;
}

Bentley โœ…
List<Integer> redIndex = new ArrayList<>();
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == 'R') redIndex.add(i);
        }
        long res = 0;
        int mid = redIndex.size() / 2; 
        for (int i = 0; i < redIndex.size(); i++) {
            res += Math.abs(redIndex.get(mid) - redIndex.get(i)) - Math.abs(mid - i);
            if (res > 1e9) {
                return -1;
            }
        }
       
        return (res > 1e9) ? -1 : (int) res;
    }

Bentley โœ…
๐—–๐—ฆ ๐—”๐—น๐—ด๐—ผ ๐Ÿ’ป ๐ŸŒ ใ€Ž๐—–๐—ผ๐—บ๐—ฝ๐—ฒ๐˜๐—ถ๐˜๐—ถ๐˜ƒ๐—ฒ ๐—ฃ๐—ฟ๐—ผ๐—ด๐—ฟ๐—ฎ๐—บ๐—บ๐—ถ๐—ป๐—ดใ€
#include <bits/stdc++.h> using namespace std; int solve(string &str) {     int ans = 0;     int j = str.size() - 1;     while (j >= 0)     {         if (str[j] == '1')         {             ans += 2;         }         else         {             ans += 1;โ€ฆ
def remove_leading_zero(xx: str) -> str:
    letter = list(xx)
    for i in range(len(letter)):
        if letter[i] == '1':
            break
        letter[i] = ' '
    ans = ''.join(letter)
    return ans.strip()

def solution(S: str) -> int:
    xy = remove_leading_zero(S)
    zero_count = 0
    one_count = 0
    f = len(xy)
    for i in range(f):
        if xy[i] == '0':
            zero_count += 1
        else:
            one_count += 1
    return zero_count + (one_count - 1) * 2 + 1

Bentley โœ…
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1) JavaScript Fundamentals
2) SQL
3) Git and GitHub
4) HTML and CSS
5) Basic projects

This are the 5 courses which are available for free, and when you will join corporate this are the basic requirements.

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Limited time ke liye hai, enroll as soon as possible (Isse pehle paid krde) ๐Ÿคž๐Ÿคž