It is the hackerearth problem setting contest.

I just realized I was wrong in my approach. I wonder why I have upvotes lol. A number can be divided by two multiple times. For example 8 has 4 possible states {8,4,2,1}, So only 2 states thing was wrong. Every number can be be divided by 2, 0 to $$$log(n)$$$ times.

There must be some solution of this using set/multiset or an appropriate data structure. But without them it can also be solved.

Let's say we are given array a[n]. We will first make all the odd numbers even by multiplying by 2. Now, we will never multiply and only divide by 2 , all numbers are their highest states. Let's find minimum of the array. Let us represent it by min. For all numbers other than min, if they are even and dividing them by 2 won't change the min, we should divide them then. Because for the fixed min, it can only decrease max. Once we do this process, let's sort the array in descending order.

Let's have 2 pointers mx and mn, representing indexes of max and min element respectively. Initially mx=0 and mn=n-1, So, now we will iterate cyclically in the array from i=0 (at anytime i will represent mx and (i+n-1)%n will represent mn. We will update our answer at each step. So, if we will divide a[i] by 2, it will become new mn. and next i will become mx and if we can't divide it by 2 (it is odd), we will stop because if we can't decrease max element, no point of decreasing other elements.

//  Copyright © 2020 Kavish Lodha. All rights reserved.
#include <bits/stdc++.h>
using namespace std;
#define vll vector<long long>
#define vvll vector<vector<long long>>
#define mll map<long long,long long>
#define max(a,b) ((a>b)?a:b)
#define min(a,b) ((a>b)?b:a)
#define pb push_back
#define prll pair<long long,long long>
typedef long long ll;
typedef long double ld;
#define ff first
#define ss second
#define rep(i,a,n)   for(ll i=a; i<n; i++)
#define rrep(i,a,n)  for(ll i=n-1; i>=a; i--)
const ll mod=1e9 + 7;
int main(){
   ios_base::sync_with_stdio(false); 
   cin.tie(NULL);
   cout.tie(NULL);
   cout<<fixed<<setprecision(20);
   ll n;
   cin>>n;
   ll a[n];
   rep(i,0,n) cin>>a[i];
   rep(i,0,n) if(a[i]%2) a[i]*=2;
   sort(a,a+n);
   rep(i,1,n){
       while(a[i]%2==0&&a[i]>=2*a[0]) a[i]/=2;
   }
   sort(a,a+n,greater<ll>());
   ll ans=a[0]-a[n-1];
   ll mx=0,mn=n-1;
   while(1){
       ans=min(ans,a[mx]-a[mn]);
       if(a[mx]%2==0){
           a[mx]/=2;
           mx=(mx+1)%n;
           mn=(mn+1)%n;
       }
       else break; 
   }
   cout<<ans;
   return 0;
}