long long cal(vector<int>&arr, long long val)
{
  long long n = arr.size();
  vector<long long>diff;
  for (auto it : arr) {
    long long dif = val - it;
    if (dif != 0) {
      diff.push_back(dif);
    }
  }
  long long cnttwo = 0;
  for (long long i = 0; i < diff.size(); ++i)
  {
    cnttwo += (diff[i] / 2);
    diff[i] %= 2;
  }

  long long cntone = 0;
  for (long long i = 0; i < diff.size(); ++i)
  {
    cntone += diff[i];
  }
  long long res = max(cntone * 2 - 1, cnttwo * 2);
  while (cnttwo > 0 && cntone  < cnttwo ) {
    cnttwo -= 1;
    cntone += 2;
    res = min(res, max(cntone * 2 - 1, cnttwo * 2));
  }
  res = min(res, max(cntone * 2 - 1, cnttwo * 2));
  return res;
}

long long findMinGeneration(vector<int>layer)
{
  long long n = layer.size();
  long long mx = *max_element(layer.begin(), layer.end());

  long long val1 = cal(layer, mx);
  long long val2 = cal(layer, mx + 1);
  long long val3 = cal(layer, mx + 2);
  long long res = min({val1, val2, val3});
  return res;
}

FINDGENERATIONS UBS 13/15 :)

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

typedef long long ll;

const int M=998244353;
ll modx(ll x){
return ((x%M + M)%M);
}
ll add(ll a, ll b){
return modx(modx(a)+modx(b));
}
ll mul(ll a, ll b){
return modx(modx(a)*modx(b));
}

ll modPow(ll a, ll b){
    if(b==0)
    return 1LL;
    if(b==1)
    return a%M;
    ll res=1;
    while(b){
        if(b%2==1)
         res=mul(res,a);
         a=mul(a,a);
         b=b/2;
        }
    return res;
}
int main()
{

int n;
cin>>n;
vector<int>v(n);
map<ll,ll>mpp;
map<ll,ll>freq;
for(int i = 0;i<n;i++){
    cin>>v[i]; 
    freq[v[i]]++;
}

ll m = 1;
ll a = 0;
for(int i = 0;i<=n;i++){
    
    a = add(a,freq[0]);
    ll y = modPow(2,n-a);
    ll mex = mul(m,y);
    mpp[i] = mex;
    m = mul(m,(modPow(2,freq[i])-1)); 
}
   
int l,r;
cin>>l>>r;

if(l>n){
    cout<<0<<endl;
    return 0;
}
else if(r>n){
   r = n;
}

int s = 0;
for(int i = l;i<=r;i++) s+=mpp[i];
cout<<s<<endl;
return 0;
}

countsubsequences 
UBS ✅

📌Brillio is hiring for ML Engineer

Expected Salary: 8-12 LPA

Apply here: https://jobs.lever.co/brillio-2/390c2eaf-f820-461d-b119-388c0078afa5/

Company Name: Aveva
Role: Software Developer

Batch eligible: 2025 grads
Apply: https://aveva.wd3.myworkdayjobs.com/AVEVA_careers/job/Hyderabad-India/Software-Developer-Graduate--India_R008700

def conversionPermutation(n):
    if n == 0:
        return "0"

    min_rep = None
    digits = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"

    for b in range(2, min(n + 1, 37)):
        rep = ""
        temp = n
        while temp > 0:
            rep = digits[temp % b] + rep
            temp //= b
        
        if min_rep is None or rep < min_rep:
            min_rep = rep

    return min_rep

Conversion Permutation ✅

INDMoney Hiring Interns
https://www.linkedin.com/posts/mourya-venkat-937264135_hiring-alert-were-looking-for-an-intern-activity-7229414800886226945-Sa3R

Sri Sri Tatva hiring for multiple Internship and Full Time roles.

https://docs.google.com/forms/d/e/1FAIpQLScRFpylPRal2_YCqbCbUobeLanNSnECHwWC_3umHwRXrl9IZA/viewform

#include <bits/stdc++.h>
using namespace std;
int main() {
    int N;
    cin >> N;

    vector<int> array(N);
    long long sum = 0;

    for (int i = 0; i < N; ++i) {
        std::cin >> array[i];
        sum += array[i];
    }
    if (sum % N == 0) {
       cout << "Yes\n";
    } else {
        int remainder = sum % N;
        cout << "No " << remainder << '\n';
    }

    return 0;
}

Wabtec ✅

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vll;

ll cost(ll r, ll c, ll j) {
    return pow(r, 2) + j * c;
}

int main() {
    ll l, c, n;
    cin >> l >> c >> n;

    vll v(n);
    for (ll i = 0; i < n; ++i) {
        cin >> v[i];
    }

    v.push_back(0);
    v.push_back(l);

    sort(v.begin(), v.end());

    ll mini = LLONG_MAX;
    ll best_jumps = 0;
    ll best_range = 0;

    for (ll i = 1; i < v.size(); ++i) {
        ll r = v[i] - v[i - 1];

        ll last_pos = 0;
        ll j = 0;
        bool f = true;

        for (ll k = 1; k < v.size(); ++k) {
            if (v[k] - last_pos > r) {
                last_pos = v[k - 1];
                j++;
                if (v[k] - last_pos > r) {
                    f = false;
                    break;
                }
            }
        }

        if (f) {
            j++;

            ll total = cost(r, c, j);

            if (total < mini) {
                mini = total;
                best_jumps = j;
                best_range = r;
            }
        }
    }

    cout << mini << " " << best_jumps << " " << best_range << endl;

    return 0;
}

Wabtec ✅

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int getMaximumPower(vector<int> arr, vector<int> power) {
    const long MOD = 1e9 + 7;
    int n = arr.size();
    int k = power.size();
    
    vector<long long> prefix_sum(n + 1, 0);
    for (int i = 1; i <= n; ++i) {
        prefix_sum[i] = (prefix_sum[i - 1] + arr[i - 1]) % MOD;
    }
    
    sort(power.begin(), power.end());
    
    long long ans = 0;
    int i = 0;
    int j = k - 1;
    
    while (i < j) {
        if (power[i] == 0) {
            ans = (ans + prefix_sum[power[j] + 1]) % MOD;
        } else {
            ans = (ans + (prefix_sum[power[j] + 1] - prefix_sum[power[i]]) % MOD + MOD) % MOD;
        }
        i++;
        j--;
    }
    
    return ans;
}

Expedia ✅

#include <vector>
using namespace std;


vector<vector<int>> computePrefixSum(const vector<vector<int>>& a) {
    int n = a.size();
    int m = a[0].size();
    vector<vector<int>> prefix(n, vector<int>(m, 0));

    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            prefix[i][j] = a[i][j];
            if (i > 0) prefix[i][j] += prefix[i-1][j];
            if (j > 0) prefix[i][j] += prefix[i][j-1];
            if (i > 0 && j > 0) prefix[i][j] -= prefix[i-1][j-1];
        }
    }

    return prefix;
}


vector<vector<int>> findMatrix(const vector<vector<int>>& a) {
    vector<vector<int>> prefix = computePrefixSum(a);
    int n = a.size();
    int m = a[0].size();
    vector<vector<int>> b(n, vector<int>(m, 0));

    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            b[i][j] = prefix[i][j];
        }
    }

    return b;
}

Meesho ✅

int getMinMachines(vector<int> start,vector<int> end){
    int n = start.size();
    sort(start.begin(),start.end());
    sort(end.begin(),end.end());
    int i = 0 , j = 0;
    int maxi = 0;
    int curr = 0;
    while(i < n && j < n){
        if(start[i] <= end[j]) i++ , curr++;
        else j++ , curr--;
        maxi = max(curr,maxi);
    }
    return maxi;
}

Expedia ✅

Ocrolus East Hiring Data Verifier ❤️❤️❤️

Job Role: Data Verifier

Qualifications: Any Graduate

Experience: 0-5 years

Job Type: Full Time

Location: Noida

Salary: 3 to 5 LPA

Skills/Requirements:
1. Review Documents of different formats and quality
2. Correct or Edit any incorrect details captured by the Ocrolus proprietary systems
3. Perform reconciliation of various amounts captured from reviewed documents to ensure high accuracy and identify discrepancies
4. Basic Understanding of financial documents and Terminologies.
5. Understanding basic accounting concepts would be an added advantage.

Date of offline Interview: 20th August, 2024

Time : 11.00 AM – 2.00 PM

Address: Ocrolus East Private Limited, A-61, 2nd Floor, ASF Building, Sector 63, Noida- 201301