PROBLEM LINK:
Practice
Contest: Division 1
Contest: Division 2
Contest: Division 3
Contest: Division 4
Author: Utkarsh Gupta
Testers: Abhinav Sharma, Venkata Nikhil Medam
Editorialist: Nishank Suresh
DIFFICULTY:
1408
PREREQUISITES:
Observation
PROBLEM:
Given N, construct a permutation of \{1, 2, 3, \ldots, N\} such that no two adjacent prefixes have the same median.
Note that the median of a set of size 2K is its K-th element in sorted order.
EXPLANATION:
We can make the following observations:
- Suppose A is a set of odd size and median M. If we add another element, say x, to A:
- If x \geq M, the median of A\cup \{x\} still remains M
- If x \lt M, the median of A\cup \{x\} is not M
- Suppose A is a set of even size and median M. If we add x to A:
- If x \leq M, the median of the new set is still M
- If x \gt M, the median of the new set is not M
This tells us the following, if we have fixed the first i elements of the permutation:
- If i is odd, the i+1-th element must be smaller than the current median
- If i is even, the i+1-th element must be larger than the current median
One construction that satisfies the above properties is as follows:
Let K = \left\lfloor \frac{N}{2} \right\rfloor. Then,
- Place K+1, K+2, \ldots, N in the odd positions
- Place K, K-1, \ldots, 1 in the even positions
For example, if N = 11 (and so K = 5), the permutation looks like [6, 5, 7, 4, 8, 3, 9, 2, 10, 1, 11].
TIME COMPLEXITY:
\mathcal{O}(N) per test case.
CODE:
Setter (C++)
//Utkarsh.25dec
#include <bits/stdc++.h>
#define ll long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define vl vector <ll>
#define all(c) (c).begin(),(c).end()
using namespace std;
ll power(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll modInverse(ll a){return power(a,mod-2);}
const int N=500023;
bool vis[N];
vector <int> adj[N];
long long readInt(long long l,long long r,char endd){
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true){
char g=getchar();
if(g=='-'){
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g && g<='9'){
x*=10;
x+=g-'0';
if(cnt==0){
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);
assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd){
if(is_neg){
x= -x;
}
if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(1 == 0);
}
return x;
} else {
assert(false);
}
}
}
string readString(int l,int r,char endd){
string ret="";
int cnt=0;
while(true){
char g=getchar();
assert(g!=-1);
if(g==endd){
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt && cnt<=r);
return ret;
}
long long readIntSp(long long l,long long r){
return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
return readString(l,r,'\n');
}
string readStringSp(int l,int r){
return readString(l,r,' ');
}
int sumN=0;
void solve()
{
int N=readInt(2,1000,'\n');
sumN+=N;
assert(sumN<=1000);
int low=1,high=N;
for(int i=1;i<=N;i++)
{
if(i%2==1)
cout<<(high--)<<' ';
else
cout<<(low++)<<' ';
}
cout<<'\n';
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
ios_base::sync_with_stdio(false);
cin.tie(NULL),cout.tie(NULL);
int T=readInt(1,30,'\n');
while(T--)
solve();
assert(getchar()==-1);
cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}
Tester (nikhil_medam, C++)
// Tester: Nikhil_Medam
#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
int t, n;
int32_t main() {
cin >> t;
while(t--) {
cin >> n;
int small = 1, large = n;
for(int i = 1; i <= n; i++) {
if(i & 1) {
cout << large-- << " ";
}
else {
cout << small++ << " ";
}
}
cout << endl;
}
return 0;
}
Editorialist (Python)
for _ in range(int(input())):
n = int(input())
a = []
L, R = 1, 2
for i in range(n):
if i%2 == 1:
# smaller
a.append(L)
L -= 1
else:
# Larger
a.append(R)
R += 1
m = min(a)
print(' '.join(str(x-m+1) for x in a))