Question

APPLE | OA | SDE1 | Coding Questions and Solutions | 2022 | Set-4

Answer 1

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#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
// using namespace chrono;
using namespace __gnu_pbds;
#define pb push_back
#define inf (1LL << 60)
#define all(x) (x).begin(), (x).end()
#define prDouble(x) cout << fixed << setprecision(10) << x
#define triplet pair<int, pair<int, int>>
#define fast_io                       \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL)
#define int long long
#define max3(a, b, c) max(max(a, b), c)
#define min3(a, b, c) min(min(a, b), c)
#define max4(a, b, c, d) max(a, max3(b, c, d))
#define min4(a, b, c, d) min(a, min3(b, c, d))
#define max5(a, b, c, d, e) max(max4(a, b, c, d), e)
#define min5(a, b, c, d, e) min(min4(a, b, c, d), e)
#define MAXN 100005
#define INF 1e18
#define mod 1000000007
#define mod2 998244353
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; // find_by_order, order_of_key
const int N = 1e5 + 10;
vector<int> adj[N];
int vis[N];
int res[N];
void dfs(int node, int par, vector<pair<int, int>> &temp, vector<int> &a)
{
    vis[node] = 1;
    temp.push_back({a[node], node});
    for (int ch : adj[node])
    {
        if (ch != par)
        {
            dfs(ch, node, temp, a);
            temp.pop_back();
        }
    }
    // sort(temp.begin(),temp.end());
    for (int i = 0; i < temp.size(); i++)
    {
        if (temp[i].first > a[node])
        {
            res[node] += 1;
        }
    }
    res[node] = res[node] * a[node];
    for (int i = 0; i < temp.size(); i++)
    {
        if (temp[i].first < a[node])
        {
            res[node] += temp[i].second;
        }
    }
    return;
}
int32_t main()
{
    fast_io;
    int tc = 1;
    // cin >> tc;
    while (tc--)
    {
        int n;
        cin >> n;
        vector<int> a(n + 1);
        for (int i = 1; i <= n; i++)
        {
            cin >> a[i];
        }
        for (int i = 1; i <= n - 1; i++)
        {
            int u, v;
            cin >> u >> v;
            adj[u].push_back(v);
            adj[v].push_back(u);
        }
        vector<pair<int, int>> temp;
        dfs(1, -1, temp, a);
        for (int i = 1; i <= n; i++)
        {
            cout << res[i] << " ";
        }
        cout << endl;
    }

    return 0;
}
Is this optimal??