#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
pair<ll, ll> dfs(int curr, int prv, const vector<vector<ll>>& adj, const vector<ll>& connect_val, int k) {
    ll tmp = connect_val[curr - 1];
    pair<ll, ll> ans = {tmp, tmp};

    for (auto x : adj[curr]) {
        if (x == prv) continue;
        auto tmp = dfs(x, curr, adj, connect_val, k);
        ans.second += tmp.second;
        ans.first += tmp.first;
    }

    ans.first = max(ans.first, -static_cast<ll>(k)); 

    return ans;
}

ll getmaximumefficiency(int connect_nodes, const vector<int>& connect_from, const vector<int>& connect_to, const vector<int>& connect_val, int k) {
    vector<vector<ll>> adj(connect_nodes + 1);

    for (int i = 0; i < connect_from.size(); i++) {
        ll u = connect_from[i], v = connect_to[i];
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    vector<ll> connect_val_ll(connect_val.begin(), connect_val.end());

    return dfs(1, 0, adj, connect_val_ll, k).first;
}

Gamekraft ✅

def lcs_length(a, b):
    m, n = len(a), len(b)
    dp = [[0] * (n + 1) for _ in range(m + 1)]
   
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if a[i - 1] == b[j - 1]:
                dp[i][j] = dp[i - 1][j - 1] + 1
            else:
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
   
    return dp[m][n]

def min_steps_to_transform(current, desired):
    L = lcs_length(current, desired)
    n = len(current)
    min_steps = n - L
   
    return min_steps

UBS✅

Company Name: Bureau

Batch eligible: 2025 and 2026 grads

Role: Engineering Intern
Apply: https://jobs.bureau.id/?ashby_jid=7d547bf9-e182-4fbc-b026-d92093fe757a


Role: DevOps Intern
Apply: https://jobs.bureau.id/?ashby_jid=fc951ad0-bdd1-4b36-b354-e3f8964aabda

LTI Mindtree Candidate got interview mail 👍🏻

def getMaxTrafficTimes(start, end):
    n = len(start)
    start.sort()
    end.sort()
    ans, cmax, j = start[0], 1, 0

    for i in range(1, n):
        while end[j] < start[i]:
            j += 1
        if i - j + 1 > cmax:
            cmax = i - j + 1
            ans = start[i]

    return ans

MSCI getmaxtraffic Time ✅

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

long triplets(long t, vector<int> d) {
    int n = d.size();
    long count = 0;

    sort(d.begin(), d.end());

    for (int i = 0; i < n - 2; ++i) {
        int j = i + 1;
        int k = n - 1;

        while (j < k) {
            int sum = d[i] + d[j] + d[k];
            if (sum <= t) {
                count += (k - j);
                ++j;
            } 
            else {
                --k;
            }
        }
    }

    return count;
}

MSCI Triplets ✅

SQL: Calendar Application Events Report 2(MSCI)✅

SELECT 
    e.dt,
    e.title,
    GROUP_CONCAT(CONCAT(g.full_name, ' <', g.email_address, '>') ORDER BY g.full_name SEPARATOR ', ') AS guests
FROM 
    events e
JOIN 
    events_guests eg ON e.id = eg.event_id
JOIN 
    guests g ON eg.guest_id = g.id
WHERE 
    g.on_vacation = 0
GROUP BY 
    e.id, e.dt, e.title
ORDER BY 
    e.dt ASC
LIMIT 5;

SELECT a.location, COUNT(f.flight_no) AS "Number of destination flights"
FROM airport a
JOIN flight f ON a.code = f.dest
GROUP BY a.location
HAVING COUNT(f.flight_no) > 0
ORDER BY COUNT(f.flight_no) DESC, a.location ASC;

AC flight sql✅