public static int numDuplicates(List<String> name, List<Integer> price, List<Integer> weight) {
    int res=0;
    HashSet<String> set = new HashSet<String>();
    for(int i=0;i<name.size();i++){
        String cmb = name.get(i) + "_" + price.get(i)+ "_" + weight.get(i);
        if(set.contains(cmb)){
            res++;
        }
        else{
            set.add(cmb);
        }
    }
    return res;
    }
}

Chubb Dupilactes products✅

#include <bits/stdc++.h>
using namespace std;
double solve(int n, const vector<int>& prices) {
    vector<int> sorted_prices = prices;
    sort(sorted_prices.begin(), sorted_prices.end(), greater<int>());
        int keep_count = (4 * n) / 5;
        long long sum = 0;
    for (int i = 0; i < keep_count; ++i) {
        sum += sorted_prices[i];
    }
        return static_cast<double>(sum) / keep_count;
}

int main() {
    int n;
    cin >> n;
    vector<int> prices(n);
    
    for (int i = 0; i < n; ++i) {
        cin >> prices[i];
    }
    
    double result = solve(n, prices);
        cout << fixed << setprecision(5) << result << endl;
    
    return 0;
}

TVS✅

int n;
    cin >> n;
    vector<int> A(n);
    for (int i = 0; i < n; i++)
    {
        cin >> A[i];
    }
    int ans = 0;
    for (int i = 0; i < n - 1; ++i) {
        while (A[i] > A[i + 1]) {
            A[i] *= 2;
            ++ans;
        }
    }
    cout << ans << endl;

\\increasing array tvs✅

def ss(D, H, Mi, S, M, L, N):
    ww = (D * 24 * 60 * 60) + (H * 60 * 60) + (Mi * 60) + S
    
    s = M * L * N
    e = ww // s
    a = ww % s
    
    r = a // (L * N)
    a = a % (L * N)
    
    d = a // N
    q = a % N
    
    return e, r, d, q
D, H, Mi, S = map(int, input().split())
M, L, N = map(int, input().split())

ww = ss(D, H, Mi, S, M, L, N)
print(f"{ww[0]} {ww[1]} {ww[2]}  {ww[3]}")

TVS✅

Company Name: Urban Company
Role: SDE 1

Batch eligible: 2024 grads

Apply: https://docs.google.com/forms/d/1-ctraB4lJb2OHVDrqM1FZB3y5Tld1E9o3IyMKyurOVI/viewform

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

const int P = 1000000007;

int solve(string N, int K) {
    if (K == 0) {
        return 1;
    }

    int len = N.size();
    int num[1010] = {0};
    int f[1010][1010][2] = {0};
    int ans = 0;

    for (int i = 2; i <= 1000; i++) {
        num[i] = num[__builtin_popcount(i)] + 1;
    }

    f[0][0][0] = 1;

    for (int i = 0; i < len; i++) {
        for (int j = 0; j <= i; j++) {
            for (int k = 0; k < 2; k++) {
                for (int l = 0; l <= (k ? 1 : (N[i] - '0')); l++) {
                    (f[i + 1][j + l][k | (l < N[i] - '0')] += f[i][j][k]) %= P;
                }
            }
        }
    }

    for (int i = 1; i <= 1000; i++) {
        if (num[i] == K - 1) {
            (ans += f[len][i][0]) %= P;
            (ans += f[len][i][1]) %= P;
        }
    }

    if (K == 1) {
        ans = (ans + P - 1) % P;
    }

    return ans;
}

int main() {
    string s;
    int k;
    cin >> s >> k;
    cout << solve(s, k) << endl;
    return 0;
}

Possible decryption ✅