Unofficial Editorial (C++)

Explanation

Let $\texttt{sum}$ be $\sum\limits_{i=1}^n t_i$.

Sort the array in ascending order. Kotivalo can read the books in the order $t_n, t_1, t_2, ..., t_{n-1}$, and Justiina can read them in the order $t_1, t_2, t_3, ..., t_n$.

If $\texttt{sum} - t_n < t_n$, Justiina would finish reading books $t_1, t_2, t_3, ..., t_{n-1}$ and would have to wait for Kotivalo to finish reading $t_n$. Afterwards, Justiina can read $t_n$ and Kotivalo can read $t_1, t_2, t_3, ..., t_{n-1}$. In total, this will take $2 \times t_n$ units of time.

If $\texttt{sum} - t_n \ge t_n$, Kotivalo would finish reading $t_n$ before Justiina finishes reading $t_1, t_2, t_3, ..., t_{n-1}$, so Justiina would not have to wait for Kotivalo to finish reading $t_n$. Thus, this will take $\texttt{sum}$ units of time.

Therefore, the answer is $\max(\texttt{sum}, 2 \times t_n)$.

Implementation

Time Complexity: $\mathcal O(N)$.

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

using ll = long long;

int main() {
	ll n;
	cin >> n;

	vector<ll> books(n);

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import java.io.*;
import java.util.*;

class ReadBook {
	public static void main(String[] args) {
		Kattio io = new Kattio();
		int n = io.nextInt();

		long books[] = new long[n];
		for (int i = 0; i < n; i++) { books[i] = io.nextLong(); }

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n = int(input())
books = [int(b) for b in input().split()]

longest_time = max(books)
total_time = sum(books)

print(max(total_time, longest_time * 2))