抽屉原理O(n)求最大间隙问题

/*********************************************************************
    程序名:
    版权: Joecai
    作者: Joecai
    日期: 2022-04-04 19:00
    说明:
*********************************************************************/
#include <bits/stdc++.h>
using namespace std;
#define y second
# define rep(i,be,en) for(int i=be;i<=en;i++)
# define pre(i,be,en) for(int i=be;i>=en;i--)
typedef pair<int, int> PII;
#define ll long long
#define endl "\n"
#define LOCAL
#define pb push_back
#define eb emplace_back
#define sp(i) setprecision(i)
const int N = 2e5 + 10, INF = 0x3f3f3f3f;
ll qpow(int a, int b) {
  ll ans = 1;
  while (b) {
    if (b & 1) ans = ans * a;
    a = (ll)a * a;
    b >>= 1;
  }
  return ans;
}
inline int read()
{
  int X = 0;
  bool flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    X = (X << 1) + (X << 3) + ch - '0';
    ch = getchar();
  }
  if (flag) return X;
  return ~(X - 1);
}
inline void write(int X)
{
  if (X < 0) {
    X = ~(X - 1);
    putchar('-');
  }
  if (X > 9) write(X / 10);
  putchar(X % 10 + '0');
}
struct dsu {
  private:
    vector<int> par, ran, cnt;
  public:
    dsu(int n) {
      par.resize(n, 0);
      iota(par.begin(), par.end(), 0);
      ran.resize(n, 0);
      cnt.resize(n, 1);
    }
    int find(int x) {
      if (par[x] == x)return x;
      else return par[x] = find(par[x]);
    }
    bool unite(int x, int y) {
      x = find(x), y = find(y);
      if (x == y) return false;
      if (ran[x] < ran[y]) {
        par[x] = y;
        cnt[y] += cnt[x];
      }
      else {
        par[y] = x;
        cnt[x] += cnt[y];
        if (ran[x] == ran[y])ran[x]++;
      }
      return true;
    }
    bool same(int x, int y) {
      return find(x) == find(y);
    }
    int getcnt(int x) {
      return cnt[find(x)];
    }
};
struct node
{
  double mi;
  double mx;
  int cnt;
};
node x[N];
double a[N];
int n;
void solve()
{
  cin >> n;
  double mx = -INF;
  double mi = INF;
  for (int i = 1; i <= n; i++)
  {
    cin >> a[i];
    mx = max(mx, a[i]);
    mi = min(mi, a[i]);
  }
  if (n == 2)
  {
    cout << mx - mi << endl;
    return;
  }
  for (int i = 0; i <= n; i++)
  {
    x[i].mi = INF, x[i].mx = -INF;
  }
  double gap = (mx - mi) / (n - 1.0);
  for (int i = 1; i <= n; i++)
  {
    int t = (int)(a[i] - mi) / gap;
    x[t].cnt++;
    x[t].mi = min(x[t].mi, a[i]);
    x[t].mx = max(x[t].mx, a[i]);
  }
  double ans = -INF;
  for (int i = 2; i <= n; i++)
  {
    if (!x[i].cnt)
    {
      x[i].mx = x[i - 1].mx;
      x[i].mi = x[i - 1].mx;
    }
    if (x[i].mi - x[i - 1].mx > ans)
    {
      ans = x[i].mi - x[i - 1].mx;
    }
  }
  cout << ans << endl;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  //#ifdef LOCAL
  //freopen("data.in.txt","r",stdin);
  //freopen("data.out.txt","w",stdout);
  //#endif
  int __ = 1;
  while (__--)
  {
    solve();
  }
  return 0;
}