EDU DIV2 B alternate binary search solution

#include <bits/stdc++.h>
using namespace std;
mt19937_64 RNG(chrono::steady_clock::now().time_since_epoch().count());
#define NeedForSpeed                  \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
#define int long long
#define all(x) (x).begin(), (x).end()
typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<vi> vvi;
typedef vector<pair<int, int>> vpi;
#define f first
#define s second
#define endl "\n"
const int mod = 1000000007;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
vi a;
bool check(int mid)
{
    int cnt = 0;
    int n = a.size();
    for (int i = 1; i < n; i++)
    {
        if ((a[i] - a[i - 1]) <= mid)
        {
            cnt++;
            i++;
        }
    }
    return cnt >= (n / 2);
}
void solve()
{
    // as in n = even all the consecutive elements would be covered so number of pairs = n/2
    // as in n = odd all but one would be covered, and the one can be compensated by atmost condition
    int n;
    cin >> n;
    a.resize(n);
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
    }
    sort(all(a));
    int low = 1, high = 3e18, ans = high;
    while (low <= high)
    {
        int mid = low + (high - low) / 2;
        if (check(mid))
        {
            ans = mid;
            high = mid - 1;
        }
        else
        {
            low = mid + 1;
        }
    }
    cout << ans << endl;
    return;
}

signed main()
{
    NeedForSpeed;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}