Is an O(n²) prefix/suffix optimization possible for Codeforces 863B (Kayaking)?

Problems and Contests
#1

Checklist

I have read the “How to ask for help” guidelines and attempted solving the problem on my own.

Problem Info

Kayaking (Codeforces 863B)

https://codeforces.com/contest/863/problem/B

Question

I solved the problem in O(n3)O(n^3)O(n3):

  • Sort the weights.
  • Enumerate the two people assigned to the single kayaks.
  • Pair the remaining people greedily (adjacent in sorted order) by scanning once.

This passes comfortably given the constraints.

However, while solving, I wondered whether the scan over the remaining people can be eliminated using some precomputation.

My idea was:

  • Fix the two singles (O(n2)O(n^2)O(n2) choices).
  • Precompute some prefix/suffix information (or similar) after sorting.
  • Answer the pairing cost for each choice in O(1)O(1)O(1), leading to an O(n2)O(n^2)O(n2) algorithm.

While thinking about it, I noticed that removing one person changes the pairing parity of the remaining people until the second removed person, so a straightforward prefix/suffix decomposition doesn’t seem to work.

My main question

Is an O(n2)O(n^2)O(n2) solution based on prefix/suffix (or another precomputation after sorting) actually possible?

Or is the parity shift fundamentally what prevents such a decomposition, making the O(n3)O(n^3)O(n3) scan the natural approach?

I’m not looking for the implementation or the full solution—I’m mainly trying to understand whether this optimization direction is fundamentally viable or a dead end.

What I’ve Tried

  • Observed that after sorting, adjacent pairing is optimal once the two singles are fixed.
  • Tried to decompose the total cost into prefix/middle/suffix contributions.
  • Noticed that removing one index shifts the pairing pattern until the second removed index, so the middle region isn’t independent.
  • Couldn’t find a clean precomputation that handles this.

My current code:

// Problem Link: https://codeforces.com/contest/863/problem/B

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

#define SOLUTION_PrefixSuffixArray

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;
    n <<= 1;

    vector<int> weights(n);
    for (int &weight : weights)
        cin >> weight;

    sort(weights.begin(), weights.end());

    // int len_diff = n - 1;
    // vector<int> diff(len_diff); // differnces[i] = weights[i + 1] - weights[i];
    // for (int i = 0; i < len_diff; i+= 2)
    //     diff[i] = weights[i + 1] - weights[i];

    int res = INT_MAX;

#ifdef SOLUTION_BruteForce

    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
        {
            int sum = 0;
            for (int k = 0, end_it = n - 1; k < end_it;)
                if ((k != i) && (k != j))
                    if ((k + 1 == i) || (k + 1 == j))
                        if ((k + 2 != i) && (k + 2 != j))
                        {
                            sum += weights[k + 2] - weights[k];
                            k += 3;
                        }
                        else
                        {
                            sum += weights[k + 3] - weights[k];
                            k += 4;
                        }
                    else
                    {
                        sum += weights[k + 1] - weights[k];
                        k += 2;
                    }
                else
                    k += 1;
            res = min(res, sum);
        }
#endif

#ifdef SOLUTION_BruteForce_Cleaner

    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
        {
            int sum = 0;

            // Works like search for one person then next throughout
            int first_person = -1;
            for (int k = 0; k < n; k++)
                if ((k != i) && (k != j))
                    // This is the first person
                    if (first_person == -1)
                        first_person = weights[k];
                    // This is the second person
                    else
                    {
                        sum += weights[k] - first_person;
                        first_person = -1;
                    }

            res = min(res, sum);
        }

#endif

#ifdef SOLUTION_PrefixSuffixArray

#endif


    cout << res << '\n';
    return 0;
}