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

vector<int> solve(int N, vector<int> a, int q, vector<vector<int>> queries) {
    vector<int> res;
    for(int i=0; i<q; i++) {
        int T = queries[i][0];
        int l = queries[i][1];
        int r = queries[i][2];
        int x = queries[i][3];
        if(T == 1) {
            int ans = a[l-1];
            for(int j=l; j<r; j++) {
                ans |= a[j];
            }
            res.push_back(ans);
        } else {
            for(int j=l-1; j<r; j++) {
                a[j] ^= x;
            }
        }
    }
    return res;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int N;
    cin>>N;
    vector<int> a(N);
    for(int i=0; i<N; i++) {
        cin>>a[i];
    }
    int q;
    cin>>q;
    vector<vector<int>> queries(q, vector<int>(4));
    for(int i=0; i<q; i++) {
        for(int j=0; j<4; j++) {
            cin>>queries[i][j];
        }
    }
    vector<int> res = solve(N, a, q, queries);
    for(int i=0; i<res.size(); i++) {
        cout<<res[i]<<" ";
    }
    cout<<endl;
    return 0;
}

Array bitwise operations✅

Company Name: Alakmalak Technologies

Role: Software Developer

Batch eligible: 2023 and 2024 passouts

Apply: https://bit.ly/3H8Dxf8

Company Name: Coinbase

Role: Software Engineer

Batch eligible: 2022 and 2023 passouts

Apply: https://www.coinbase.com/en-gb/careers/positions/5598961?gh_jid=5598961&gh_src=20687b321us%E2%80%A6

Saks subarray product✅

long long solve(vector<int>& nums, int k) {
    if (k <= 1) return 0;
    int n = nums.size();
    long long p = 1;
    int i = 0, j = 0;
    long long ans = 0;
    while (j < n) {
        p *= nums[j];
        while (i <= j && p > k) {
            p /= nums[i];
            i++;
        }
        ans += j - i + 1;
        j++;
    }
    return ans;
}

int solve(int N, int M, vector<int>& plates) {
    sort(plates.begin(), plates.end());
    int left = 0, right = 1e9+7;
    while (left < right) {
        int mid = left + (right - left + 1) / 2;
        int sum = 0;
        for (int i = N - 1; i >= 0 && plates[i] >= mid; i--) {
            sum += plates[i] - mid;
            if (sum >= M) break;
        }
        if (sum >= M) left = mid;
        else right = mid - 1;
    }
    return left;
}

Thanksgiving ✅

Company - HITACHI

Role - R&D Engineer

Batch - 2023 , 2024

Exp CTC - 15 lpa

Location - Bengaluru

Link - https://www.hitachienergy.com/careers/open-jobs/details/JID3-60485

substring minimization ✅

Array bitwise operations ✅

Airblack is hiring SDE Interns

For 2024/2023 Grads
Duration: 6 months

https://www.linkedin.com/posts/suyash-agrawal-25204a131_hiring-hiringdevelopers-fullstackdevelopment-activity-7150420023948234752-z3SE?utm_source=share&utm_medium=member_android

Company - IntelliFill Tech Pvt Ltd

Role - Software Engineer Internship

Batch - 2023/2024/2025

Apply Link - https://unstop.com/o/o0V4HyK?lb=cdmoBrn&utm_medium=Share&utm_source=shortUrl

vector<int> solve(vector<int>& mem, vector<vector<int>>& queries) {
    map<int, pair<int, int>> alloc;
    int id = 1;
    vector<int> res;
    for (auto& q : queries) {
        if (q[0] == 0) {
            int x = q[1];
            int start = -1;
            int count = 0;
            for (int i = 0; i < mem.size(); ++i) {
                if (mem[i] == 0) {
                    if (start == -1 && i % 8 == 0) {
                        start = i;
                    }
                    if (start != -1) {
                        ++count;
                    }
                    if (count == x) {
                        break;
                    }
                } else {
                    start = -1;
                    count = 0;
                }
            }
            if (count == x) {
                fill(mem.begin() + start, mem.begin() + start + x, 1);
                alloc[start] = {id, x};
                res.push_back(start);
                ++id;
            } else {
                res.push_back(-1);
            }
        } else {
            int eraseId = q[1];
            int length = -1;
            for(auto it = alloc.begin(); it != alloc.end(); ++it) {
                if(it->second.first == eraseId) {
                    fill(mem.begin() + it->first, mem.begin() + it->first + it->second.second, 0);
                    length = it->second.second;
                    alloc.erase(it);
                    break;
                }
            }
            res.push_back(length);
        }
    }
    return res;
}

Tradedsk ✅