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Accenture 11 am slot{ On campus}▶️

90-90 MCQ 🧑‍💻
2-2 coding -Python🧑‍💻

Using remote access ❤️

Mission accomplished 👨‍💻💫🏆

#Team_Alpha♾️

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@mrtrueliving_ix ▶️

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@mrtrueliving_ix ▶️

#Accenture #PADA #ONCampus #shortlisted
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🔸Net core

🔸Amazon

🔸MAQ

🔸Throughtworks

🔸Ibm

🔸Asian paints {igniTe}

🔸Accenture ON campus

🔸Oracle

🔸 Infosys

🔸 Cognizant aptitude & Technical

🟨All placement help available🟨

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DM -@mrtrueliving_ix☄️

DM  -@mrtrueliving_ix☄️
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MAQ Hiring for Developer role
Exp : 0 year

Apply Link : https://forms.office.com/r/eaWWtnuMx7

Join Telegram : https://t.me/code_alphix
1
#include <iostream>
#include <vector>
using namespace std;

const int MOD = 1e9 + 7;

class SegmentTree {
    int n;
    vector<long long> t;
    bool isSum;

public:
    SegmentTree(const vector<int>& v, bool sum) : n(v.size()), isSum(sum) {
        t.resize(2 * n, sum ? 0 : 1);
        for (int i = 0; i < n; ++i) {
            t[i + n] = v[i];
        }
        for (int i = n - 1; i > 0; --i) {
            t[i] = isSum ? (t[2 * i] + t[2 * i + 1]) % MOD : (t[2 * i] * t[2 * i + 1]) % MOD;
        }
    }

    long long query(int l, int r) {
        long long res = isSum ? 0 : 1;
        l += n;
        r += n;
        while (l < r) {
            if (l & 1) {
                res = isSum ? (res + t[l]) % MOD : (res * t[l]) % MOD;
                ++l;
            }
            if (r & 1) {
                --r;
                res = isSum ? (res + t[r]) % MOD : (res * t[r]) % MOD;
            }
            l /= 2;
            r /= 2;
        }
        return res;
    }
};

vector<long long> HardQueries(int n, const vector<int>& v, int q, const vector<vector<int>>& queries) {
    vector<long long> ans;
    SegmentTree sumTree(v, true);
    SegmentTree prodTree(v, false);

    for (const auto& x : queries) {
        int type = x[0], a = x[1] - 1, b = x[2];
        vector<int> indices;
        for (int i = a; i < n; i += b) {
            indices.push_back(i);
        }

        if (type == 0) {
            long long res = 0;
            for (int idx : indices) {
                res = (res + sumTree.query(idx, idx + 1)) % MOD;
            }
            ans.push_back(res);
        } else if (type == 1) {
            long long res = 1;
            for (int idx : indices) {
                res = (res * prodTree.query(idx, idx + 1)) % MOD;
            }
            ans.push_back(res);
        }
    }

    return ans;
}


Queries and fruits // OYO👍
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def solve(n, m, arr):
    mod = 10**9 + 7
    fact = [1] * (n + 1)
    for i in range(2, n + 1):
        fact[i] = fact[i - 1] * i % mod

    arr.sort()
    if n == 1:
        return 1

    val = arr.pop()
    arr[-1] += val

    def mod_inverse(a, mod):
        return pow(a, mod - 2, mod)
   
    ans = fact[n]
    temp = 1
    for count in arr:
        temp = (temp * fact[count]) % mod
   
    temp = mod_inverse(temp, mod)
    return (ans * temp) % mod

minimum arrangements // Oyo
#include <iostream>
#include <vector>
#include <algorithm>

const int MOD = 1e9 + 7;

using namespace std;

long long solve(const vector<int>& a, int b) {
    int c = a.size();
    long long d = 0;
    for (int i = 0; i < c; i++) {
        d += a[i];
    }

    vector<long long> e(b + 1, 0);
    e[0] = 1;

    for (int i = 0; i < c; i++) {
        for (int j = b; j >= a[i]; j--) {
            e[j] = (e[j] + e[j - a[i]]) % MOD;
        }
    }
    long long f = 1;
    for (int i = 0; i < c; i++) {
        f = (f * 2) % MOD;
    }

    long long g = 0;
    for (int i = 0; i <= b; i++) {
        g = (g + e[i]) % MOD;
    }

    long long h = (f - 2 * g + MOD) % MOD;
    return h;
}

Sum of the elements // OYO
Company Name: Optum
Role: Software Engineer
Batch eligible: 2023 and 2024 grads

Apply: https://careers.unitedhealthgroup.com/job/21243748/software-engineer-hyderabad-in
IBM

Both Coding
📱

Test accomplished
📣

DM for any placement help

-
@mrtrueliving_ix

-
@mrtrueliving_ix

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@mrtrueliving_ix

All placement help available
#IBM #coding  #Done
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1
Swiggy BACKEND ASDE 

Test accomplished

Coding+ SQL + all MCQs ▶️

All placement help available

OA HELP: @mrtrueliving_ix💯

#Swiggy #BE #ASDE #ALL-DONE ☺️
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Netcore
Both codes are done 📱
Mission accomplished 📱
All placement help available
@mrtrueliving_ix

▪️ Company Name: Netcore

▪️ Role : Backend Developer

▪️0-5 years experience

▪️ Any Graduates | Any Batch
Apply Línk
https://assessment.hackerearth.com/challenges/new/hiring/netcore-backend-developer-hiring-challenge-november-2024-round-1/

Direct test


Apply and Contact For Help
@mrtrueliving_ix
@mrtrueliving_ix
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All OA help available
☄️ swiggy { FE//BE}
☄️Net core
☄️Amazon
☄️MAQ
☄️Throughtworks (tomorrow end)
☄️Ibm
☄️Asian paints {igniTe}
☄️Accenture ON campus
☄️Oracle
☄️Infosys
☄️Cognizant aptitude & Technical

🟨All placement help available🟨

🟫Remote access { 💯} Success rate { 💯}🟫

All OA help { ON campus or Off campus}


DM  -@mrtrueliving_ix☄️

DM  -@mrtrueliving_ix☄️
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Tata 1g BOTH codes are done 📣

DM @mrtrueliving_ix ❤️▶️
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Netcore👍

Both codes are done
📱

💯 Plag free
📱

Mission accomplished
👑

OA HELP

@MRTRUELIVING_IX

@MRTRUELIVING_IX

💯 TEST CLEARANCE
💙

#NetCore #SD #Coding #round 1
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