LAYERCAKE - Editorial

editorial
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1

PROBLEM LINK:

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

Author: raysh07
Tester: sushil2006
Editorialist: raysh07

DIFFICULTY:

Cakewalk

PREREQUISITES:

None

PROBLEM:

Chef has N options for first layer, A_1, A_2, \ldots, A_N, and M options for second layer, B_1, B_2, \ldots, B_M. A cake is valid when the first layer has a strictly larger size than the second layer. How many cakes can Chef make?

EXPLANATION:

We can simply run a nested for-loop over all options of first layer (say i), and all options of second layer (say j), and then add 1 to answer if A_i > B_j.

The inner for loop will run M times, and the outer for loop will run N times, giving a total time complexity of O(N \cdot M).

It is also possible to solve the problem in O(N + M) by sorting the arrays and then using 2 pointers, but it was unnecessary.

TIME COMPLEXITY:

\mathcal{O}(N \cdot M) per testcase.

CODE:

Editorialist'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, m; cin >> n >> m;
    
    vector <int> a(n), b(m);
    for (auto &x : a) cin >> x;
    for (auto &x : b) cin >> x;
    
    int ans = 0;
    
    for (int i = 0; i < n; i++){
        for (int j = 0; j < m; j++) if (a[i] > b[j]){
            ans++;
        }
    }
    
    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;
}```