BUYGAME - Editorial

editorial
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PROBLEM LINK:

Practice
Contest: Division 1
Contest: Division 2
Contest: Division 3
Contest: Division 4

Author: raysh07
Tester: sushil2006
Editorialist: iceknight1093

DIFFICULTY:

Simple

PREREQUISITES:

None

PROBLEM:

There are N items. The i-th item costs A_i if you buy it on any day other than i, and B_i if you buy it on day i.
You can buy one item every day. Find the minimum total cost of buying all N items.

EXPLANATION:

The absolute best we can do is to buy item i at a cost of \min(A_i, B_i).
This means:

  • If A_i \lt B_i, we want to buy item i on some day other than i, but any of the other N-1 days are equally valid.
  • if A_i \gt B_i, we want to buy item i on exactly day i.
  • If A_i = B_i, it doesn’t matter.

First, let’s buy every item with A_i \gt B_i on its respective day.
This will leave us with some items remaining - suppose there are k such items, indexed i_1, i_2, \ldots, i_k.

Now,

  • If k = 0, we’re done: every item has been bought for its minimum possible price.
  • If k \gt 1, we can always buy every remaining item i_j on a day other than i_j itself.
    For example, buy item i_1 on day i_2, item i_2 on day i_3, …, item i_k on day i_1.

That leaves us with the k = 1 case, meaning there’s exactly one item we want to buy on a different day.
However, if we buy item i_1 on a day d that is different from i_1, item d cannot also be bought on day d - and so needs to be bought on some other day.
Since we’re in the case where it’s optimal to buy every item other than i_1 on its own day, we might as well buy item d on day i_1 and leave everything else alone.

If item i_1 is bought on day d, the total cost is A_{i_1} + A_d, plus the sum of B_i values of all other indices.
Try every d and take the best among them - to quickly find the sum of B_i values of all but these two indices, store the sum of all the B_i and subtract two values from it instead.

Apart from that, we can also choose to just buy item i_1 on day i_1, for a cost of B_{i_1} (and then everything else is also bought for B_i since they don’t need to be moved to a different day anymore).

So,

  1. If k \neq 1, the answer is simply \sum_{i=1}^N \min(A_i, B_i).
  2. If k = 1, the answer is the minimum of \sum_{i=1}^N B_i, and the minimum of buying i_1 on day d across all d\neq i which can be found by iterating over d.

TIME COMPLEXITY:

\mathcal{O}(N) per testcase.

CODE:

Author's code (C++) 
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define INF (int)1e18

mt19937_64 RNG(chrono::steady_clock::now().time_since_epoch().count());

void Solve() 
{
    int n; cin >> n;
    
    vector <int> a(n), b(n);
    for (auto &x : a) cin >> x;
    for (auto &x : b) cin >> x;
    
    int ans = 0;
    for (auto x : b){
        ans += x;
    }
    
    for (int i = 0; i < n; i++){
        a[i] -= b[i];
    }
    
    sort(a.begin(), a.end());
    int val = 0;
    int curr = 0;
    int sum = ans;
    
    for (int i = 0; i < n; i++){
        val += a[i];
        if (i){
            ans = min(ans, sum + val);
        }
    }
    
    cout << ans << "\n";
}

int32_t main() 
{
    auto begin = std::chrono::high_resolution_clock::now();
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    // freopen("in",  "r", stdin);
    // freopen("out", "w", stdout);
    
    cin >> t;
    for(int i = 1; i <= t; i++) 
    {
        //cout << "Case #" << i << ": ";
        Solve();
    }
    auto end = std::chrono::high_resolution_clock::now();
    auto elapsed = std::chrono::duration_cast<std::chrono::nanoseconds>(end - begin);
    cerr << "Time measured: " << elapsed.count() * 1e-9 << " seconds.\n"; 
    return 0;
}
Editorialist's code (PyPy3) 
for _ in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    
    bad = []
    for i in range(n):
        if b[i] > a[i]: bad.append(i)
    
    if len(bad) != 1:
        ans = sum(min(a[i], b[i]) for i in range(n))
    else:
        ans = init = sum(b)
        i = bad[0]
        for j in range(n):
            if i == j: continue
            ans = min(ans, init - b[i] - b[j] + a[i] + a[j])
    print(ans)
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