HackerEarth Easy Strong Permutation problem solution

In this HackerEarth Easy Strong, Permutation problem solution Kevin has a sequence of integers a1, a2, …, an. Define the strength of the sequence to be

|a1 – a2| + |a2 – a3| + … + |an-1 – an| + |an – a1|.

Kevin wants to make his sequence stronger, so he reorders his sequence into a new sequence b1, b2, …, bn. He wants this new sequence to be as strong as possible. What is the largest possible strength of the resulting sequence?

HackerEarth Easy Strong Permutation problem solution.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll p = 1e9+7;
const int MAXN = 1e5+5;

ll nums[MAXN], n, sum;

int main() {
    cin >> n; for (int i = 0; i < n; ++i)
        cin >> nums[i];

    sort(nums, nums+n);
    for (int i = 0; i < n/2; ++i)
        sum -= 2*nums[i], sum += 2*nums[n-1-i];

    cout << sum << "n";
}

Second solution

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define MAXN 100005
int arr [MAXN]; 
main(){
    ios_base::sync_with_stdio(0); 
    int n; cin >> n; assert(1<=n && n<=1e5); for (int g=1; g<=n; g++) {cin >> arr[g]; assert(0<=abs(arr[g]) && abs(arr[g])<=1e9);} 
    sort(arr+1, arr+n+1); 
    vector <int> a, b; 
    for (int g=1; g<=n/2; g++) a.push_back(arr[g]); 
    for (int g=n/2+1; g<=n; g++) b.push_back(arr[g]); 
    int ans = 0; 
    vector <int> c; //
    if (a.size() >= b.size()){
        int ans1 = 0; 
        c.clear(); 
        for (int g=0; g<b.size(); g++) {c.push_back(a[g]); c.push_back(b[g]);}
        if (a.size() > b.size()) c.push_back(a.back()); 
        for (int g=0; g<c.size()-1; g++) ans1+=abs(c[g+1] - c[g]); 
        ans1+=abs(c.back() - c[0]);
        ans = max(ans, ans1); 
    }
    else{
        int ans2 = 0; 
        c.clear(); 
        for (int g=0; g<a.size(); g++){
            c.push_back(b[g]); 
            c.push_back(a[g]);
        }
        if (b.size() > a.size()) c.push_back(b.back()); 
        for (int g=0; g<c.size()-1; g++) ans2+=abs(c[g+1] - c[g]); 
        ans2+=abs(c.back() - c[0]);
        ans = max(ans, ans2); 
    }//
    cout << ans;
    return 0; 
}

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