## Electronics Shop

Solutions A simple solution consists in sorting both sequences and running two nested for loops to calculate the possible costs. Although this solution is \(O((n+m) \cdot log(n+m)) + n \cdot m)\), it’s acceptable since the input is small:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 int s, n, m; cin >> s >> n >> m; vector<int> k(n); copy_n(istream_iterator<int>(cin), n, begin(k)); vector<int> u(m); copy_n(istream_iterator<int>(cin), m, begin(u)); sort(begin(k), end(k), greater<>{}); sort(begin(u), end(u), greater<>{}); auto cost = -1; for(auto i : k) { for (auto j : u) { if (i+j<=s) cost = max(cost, i+j); } } cout << cost; Linq-based solution, with cartesian product (by fabrizio_sanmar1):
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