function getSumOfFactors(num) {
    let sum = 0;
    for (let i = 1; i <= Math.sqrt(num); i++) {
        if (num % i === 0) {
            sum += i;
            if (i !== num / i) {
                sum += num / i;
            }
        }
    }
    return sum;
}

function maxSubsetSum(arr) {
    return arr.map(num => getSumOfFactors(num));
}

Factset
Largest Subset Sum ✅

Who have Google and Goldman Sachs exam guys?
Send questions here solutions will be provide

#include <iostream>
#include <vector>
#include <cstring>
using namespace std;
const int MAXN = 200001;
vector<int> tree[MAXN];
char C[MAXN];
int result[MAXN];
bool isPalindrome[MAXN];
bool visited[MAXN];
bool checkPalindrome(const string& s) {
    int n = s.size();
    for (int i = 0; i < n / 2; ++i) {
        if (s[i] != s[n - i - 1]) {
            return false;
        }
    }
    return true;
}
string dfs(int u) {
    visited[u] = true;
    string S = "";

    for (int v : tree[u]) {
        if (!visited[v]) {
            S += dfs(v);
        }
    }

    S += C[u];
    isPalindrome[u] = checkPalindrome(S);
    return S;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
   
    int N;
    cin >> N;

    for (int i = 1; i < N; ++i) {
        int u, v;
        cin >> u >> v;
        tree[u].push_back(v);
        tree[v].push_back(u);
    }
    for (int i = 1; i <= N; ++i) {
        cin >> C[i];
    }

    memset(visited, false, sizeof(visited));
    dfs(1);

    int Q;
    cin >> Q;
    while (Q--) {
        int u;
        cin >> u;
        cout << (isPalindrome[u] ? 1 : 0) << "\n";
    }

    return 0;
}.

// FIND PALIDROME
Google ✅

#include <iostream>
#include <vector>
using namespace std;

int main() {
    int T;
    cin >> T;

    while (T--) {
        int N;
        cin >> N;
        vector<int> weights(N);
        for (int i = 0; i < N; ++i) {
            cin >> weights[i];
        }

        int B;
        cin >> B;
        int totalWeight = 0;
        for (int i = 0; i < N; ++i) {
            totalWeight += weights[i];
        }

        int totalCapacity = B * 10;
        if (totalWeight <= totalCapacity) {
            cout << "True" << endl;
        } else {
            cout << "False" << endl;
        }
    }

    return 0;
}

Goldman Sachs ✅

#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>
using namespace std;
long clinicShed(vector<long> patients, int k) {
    sort(patients.begin(), patients.end());
 
    long minShedLength = LONG_MAX;
    int n = patients.size();
        for (int start = 0; start <= n - k; ++start) {
        int end = start + k - 1;
        long length = patients[end] - patients[start] + 1;
        minShedLength = min(minShedLength, length);
    }
   
    return minShedLength;
}.

Vaccination Drive ✅
GOLDMAN

String machine try this
Google✅

#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>
using namespace std;

int f(int id, int taken, int t1, int t2, int k, vector<int> &a, vector<vector<int>> &dp) {
    if (taken == 2 * k) {
        return t1 ^ t2;
    }
    if (id >= a.size()) {
        return INT_MIN;
    }
    if (dp[id][taken] != -1) {
        return dp[id][taken];
    }
   
    int take = INT_MIN, ntake = INT_MIN;
   
    if (taken >= k) {
        take = f(id + 1, taken + 1, t1, t2|a[id], k, a, dp);
    } else {
        take = f(id + 1, taken + 1, t1|a[id],t2, k, a, dp);
    }
   
    ntake = f(id + 1, taken, t1, t2, k, a, dp);
   
    return dp[id][taken] = max(take, ntake);
}

OR XOR ✅
Google

def calculateF(B, K):
    orPart = 0
    xorPart = 0
    for i in range(K):
        orPart |= B[i]
    for i in range(K, 2 * K):
        xorPart ^= B[i]
    return orPart + xorPart

def findMaxF(N, K, A):
    maxF = 0
    from itertools import combinations
    for subset in combinations(A, 2 * K):
        currentF = calculateF(subset, K)
        maxF = max(maxF, currentF)
    return maxF

def main():
    T = int(input())
    for _ in range(T):
        N, K = map(int, input().split())
        A = list(map(int, input().split()))
        print(findMaxF(N, K, A))

if name == "main":
    main()                  

// COMPLEX ✅
Google

String Machine ✅
Google

Range Function ✅
Google

using ll = long long;

class SegTree {
    vector<ll> tree;
    vector<ll> A;
    ll n;

public:
    SegTree(const vector<int>& arr) {
        n = arr.size();
        A = vector<ll>(arr.begin(), arr.end());
        tree.resize(4 * n, 0);
        build(0, 0, n - 1);
    }

    void build(ll node, ll start, ll end) {
        if (start == end) {
            tree[node] = A[start];
        } else {
            ll mid = (start + end) / 2;
            build(2 * node + 1, start, mid);
            build(2 * node + 2, mid + 1, end);
            tree[node] = tree[2 * node + 1] + tree[2 * node + 2];
        }
    }

    void upd(ll idx, ll val, ll node, ll start, ll end) {
        if (start == end) {
            A[idx] = val;
            tree[node] = val;
        } else {
            ll mid = (start + end) / 2;
            if (idx <= mid) {
                upd(idx, val, 2 * node + 1, start, mid);
            } else {
                upd(idx, val, 2 * node + 2, mid + 1, end);
            }
            tree[node] = tree[2 * node + 1] + tree[2 * node + 2];
        }
    }

    ll qry(ll L, ll R, ll node, ll start, ll end) {
        if (R < start || end < L) {
            return 0;
        }
        if (L <= start && end <= R) {
            return tree[node];
        }
        ll mid = (start + end) / 2;
        ll left = qry(L, R, 2 * node + 1, start, mid);
        ll right = qry(L, R, 2 * node + 2, mid + 1, end);
        return left + right;
    }

    void upd(ll idx, ll val) {
        upd(idx, val, 0, 0, n - 1);
    }

    ll qry(ll L, ll R) {
        return qry(L, R, 0, 0, n - 1);
    }
};

vector<ll> solve(int N, int l, vector<int> A, vector<vector<int>> query) {
    SegTree segTree(A);
    vector<ll> results;

    for (const auto& q : query) {
        if (q[0] == 1) {
            int idx = q[1];
            int val = q[2];
            segTree.upd(idx - 1, val);
        } else if (q[0] == 2) {
            int L = q[1];
            int R = q[2];
            ll sum = 0;
            for (int i = L - 1; i < R; ++i) {
                for (int j = i; j < R; ++j) {
                    sum += segTree.qry(i, j);
                }
            }
            results.push_back(sum);
        }
    }

    return results;
}

Source : Hola

#include <bits/stdc++.h>
using namespace std;
#define ll long long
void dfs2(vector<ll>g[],ll v,ll p,vector<vector<ll>>&dp,vector<ll>&a,ll k)
{
    dp[v][0]=a[v];
    dp[v][1]=a[v]^k;
    for(auto it:g[v])
    {
        if(it!=p)
        {
            dfs2(g,it,v,dp,a,k);
            dp[v][0]+=dp[it][0];
            dp[v][1]+=dp[it][1];
        }
    }
}
void dfs(vector<ll>g[],ll v,ll p,vector<vector<ll>>&dp,ll&ans)
{
    ans=max(ans,dp[1][0]-dp[v][0]+dp[v][1]);
    for(auto it:g[v])
    {
        if(it!=p)
        {
            ans=max(ans,dp[1][1]-dp[it][1]+dp[it][0]);
            dfs(g,it,v,dp,ans);
        }
    }
}
ll solve(ll n,ll k,vector<ll>&a,vector<vector<ll>>&e)
{
    vector<ll>g[n+10];
    ll i;
    assert(1<=n and n<=1e5);
    for(ll i=0;i<n;i++)
    {
        ll u=e[i][0];
        ll v=e[i][1];
        g[v].push_back(u);
        g[u].push_back(v);
    }
   a.insert(a.begin(),0);
   vector<vector<ll>>dp(n+10,vector<ll>(2));
   ll ans=0;
   dfs2(g,1,0,dp,a,k);
   ans=max(ans,dp[1][0],dp[1][1]);
   dfs(g,1,0,dp,ans);
   return ans;
}

Xor Subtree ✅
Google

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

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    int n, k;
    cin >> n >> k;

    map<int, int> mp;
    for (int i = 0; i < n; ++i) {
        int l, r, x;
        cin >> l >> r >> x;
        mp[l] += x;
        mp[r + 1] -= x;
    }

    vector<int> al;
    int sum = 0;
    for (auto& e : mp) {
        sum += e.second;
        al.push_back(sum);
    }

    int ans = 0, maxi = 0;
    unordered_map<int, int> lp;
    for (int i = 0; i < al.size(); ++i) {
        int num = al[i];
        int curr = lp[num - k] + 1;
        lp[num] = max(lp[num], curr);
        if (ans < lp[num] || (ans == lp[num] && maxi > num)) {
            maxi = num;
            ans = lp[num];
        }
    }

    cout << ans << " ";
    for (int i = ans-1; i >=0; i--) {
        cout << maxi - i * k << " ";
    }
    cout << "\n";

    return 0;
}

complex subsequence ✅

#include<bits/stdc++.h>
using namespace std;
    
int findMessages (int N, vector<string> A) {
    int ans=N;
    set<string> st;

    for(int i=0; i<N; i++) {
        string s;
        for(int j=0; j<A[i].size(); j++)
            s += 'a' + ('z' - A[i][j]);
       
        if (st.find(s) != st.end())
            ans -= 1;

        st.insert(A[i]);
    }

    return ans;
}
    
int main() {
    
    ios::sync_with_stdio(0);
    cin.tie(0);
    int N;
    cin >> N;
    vector<string> A(N);
    for(int i_A = 0; i_A < N; i_A++)
    {
        cin >> A[i_A];
    }
    
    int out_;
    out_ = findMessages(N, A);
    cout << out_;
}

Alice and messages ✅

vector<string> solve(string t, int Q, vector<string> q) {
    stringstream ss(t);
    string tk;
    vector<string> p;
    while (getline(ss, tk, '\"')) {
        p.push_back(tk);
    }

    vector<string> r;
    for (string qry : q) {
        size_t pos = qry.find_last_of('.');
        qry = qry.substr(pos + 1);
        int j = 0;
        for (j = 0; j < p.size(); j++) {
            if (p[j] == qry) {
                break;
            }
        }
        r.push_back(p[j + 2]);
    }
    return r;
}

Jason praising ✅

def solve(N, A, D, V):
    info_index = {
        'bank_account_number': 0,
        'account_holder_first_name': 1,
        'account_holder_last_name': 2,
        'registered_mobile_number': 3,
        'branch_code': 4
    }
   
    filtered_accounts = [account for account in A if str(account[info_index[D]]) == V]
    sorted_accounts = sorted(filtered_accounts, key=lambda x: int(x[0]))

    output = "\n".join([" ".join(map(str, account)) for account in sorted_accounts])
    return output

N = int(input())
A = []
for _ in range(N):
    account_info = input().split()
    account_info[0] = int(account_info[0])
    account_info[3] = int(account_info[3])
    account_info[4] = int(account_info[4])
    A.append(account_info)

D = input()
V = input()

result = solve(N, A, D, V)
print(result)

Searching google pay saved account✅

Tax Payment ✅

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

int main()
{
    int n; cin >> n;
    vector<char>decode(62);
    for (int i = 0; i < 10; ++i)
    {
        decode[i] = (i + '0');
    }
    for (int i = 10; i <= 35 ; ++i)
    {
        decode[i] = ((i - 10) + 'A');
    }
    for (int i = 36; i <= 61 ; ++i)
    {
        decode[i] = ((i - 36) + 'a');
    }

    string res = "";
    while (n) {
        int temp = n % 62;
        res += decode[temp];
        n /= 62;
    }
    reverse(res.begin(), res.end());
    cout << res << endl;
}

Encode it ✅