ARRSORT - EDITORIAL

Practice

Setter: Abhinav Sharma
Testers: Nishank Suresh
Editorialist: Mithil Shah

1681

Basic Math

PROBLEM

You are given an unsorted permutation P of size N. An operation is defined as:

Swap P_i and P_{i+K} for any i in the range [1,N−K].

Find the maximum value of K, such that, the permutation P can be sorted by applying any finite number of operations.

EXPLANATION

Observation:

Assume P_i is not in i^{th} position where 1 \le i \le N, to move P_i to i^{th} position, K should be such that it is a factor of the displacement i.e. |P_i - i|.

Proof

We can say that we would need to move P_i in one direction only, as moving in the other direction would cancel out the previous one. Let us say that we take x moves to move P_i to i^{th} position. In order for this to be possible, x \times K =|P_i - i|. Hence x = |P_i - i|/K where x is a natural number. This implies K divides |P_i - i| or in other words, K is a factor of |P_i - i|.

For example, in [5,2,3,4,1], we need to move 5 from 1^{st} position to 5^{th} position, this can be done by setting K=4 or K=2 or K=1.

Let us calculate all such displacement for which P_i is not in it’s sorted position.

Now, to calculate the maximum value of K that can make the array sorted, we need to find K such that for all P_i not in i^{th} position, we can send it to i^{th} position by performing finite operations.

Hence, K must be a factor of all such displacement and should be maximum. So, we need to find maximum common factor or greatest common divisor of all such displacements. Therefore, the answer would be the gcd of all such displacements.

To calculate this, we will iterate over all N integers, calculate the gcd of abs(P_i-i) where 1 \le i \le N. This would be the required answer.

TIME COMPLEXITY

The time complexity is O(N) per testcase.

SOLUTIONS

Setter's Solution
#include <bits/stdc++.h>
using namespace std;

/*
------------------------Input Checker----------------------------------
*/

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 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){
}
long long readIntLn(long long l,long long r){
}
}
}

/*
------------------------Main code starts here----------------------------------
*/

const int MAX_T = 1e5;
const int MAX_N = 1e5;
const int MAX_SUM_LEN = 1e5;

#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define ff first
#define ss second
#define mp make_pair
#define ll long long
#define rep(i,n) for(int i=0;i<n;i++)
#define rev(i,n) for(int i=n;i>=0;i--)
#define rep_a(i,a,n) for(int i=a;i<n;i++)
#define pb push_back

ll sum_n = 0, sum_m = 0;
int max_n = 0, max_m = 0;
int yess = 0;
int nos = 0;
int total_ops = 0;
ll mod = 998244353;

using ii = pair<ll,ll>;

void solve(){
sum_n+=n;

int p[n];
rep(i,n){
}

int g = 0;
rep(i,n){
g = __gcd(g, abs((p[i]-1)-i));
}

sort(p,p+n);
rep(i,n) assert(p[i]=i+1);

if(g==0) g=-1;
cout<<g<<'\n';
}

signed main()
{

#ifndef ONLINE_JUDGE
freopen("input.txt", "r" , stdin);
freopen("output.txt", "w" , stdout);
#endif
fast;

int t = 1;

for(int i=1;i<=t;i++)
{
solve();
}

assert(getchar() == -1);
assert(sum_n<=1e5);

cerr<<"SUCCESS\n";
cerr<<"Tests : " << t << '\n';
cerr<<"Sum of lengths : " << sum_n<<'\n';
//cerr<<"Maximum answer : " << max_n <<'\n';
// // cerr<<"Total operations : " << total_ops << '\n';
// cerr<<"Answered yes : " << yess << '\n';
// cerr<<"Answered no : " << nos << '\n';

cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}

Tester's Solution
import math
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
pos = [0]*(n+1)
for i in range(n):
pos[a[i]] = i+1
ans = 0
for i in range(1, n+1):
ans = math.gcd(ans, i - pos[i])
print(ans)

Editorialist's Solution
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define vi vector<int>
#define rep(i, n) for (int i = 0; i < n; i++)
#define endl '\n'
#define fastio                    \
ios_base::sync_with_stdio(0); \
cin.tie(0);                   \
cout.tie(0);
void tosolve()
{
int n;
cin>>n;
int p[n+1];
for(int i=1;i<=n;i++) cin>>p[i];
int ans=0; //this will store gcd value
for(int i=1;i<=n;i++)
{
ans=__gcd(ans,abs(p[i]-i));
}
cout<<ans<<endl;
}

int32_t main()
{
fastio;
int t;
cin>>t;
while(t--)
{
tosolve();
}
return 0;
}


Feel free to share your approach. Suggestions are welcomed.

4 Likes

Can someone point out why my approach fails?
Approach: for every number let k denotes the distance from its proper position, then all the factors of k can also be used to place curr elem at its proper place so we mark all factor of k

then the maximum k with freq of cnt(cnt=number of elem not at right position).
IDK why it fails

1 Like

I think there should be a log factor in the time complexity because of taking gcd. Correct me if I am wrong.

2 Likes

For example, in [5,2,3,4,1], we need to move 5 from 1st
position to 5th position, this can be done by setting K=4 or K=2 or K=1.

how can it be done by setting k = 2 or k = 1
I just read it again and realized that we’re doing it multiple times. i.e 2*(2) or 4*(1)
Correct me if I’m wrong.

ans = math.gcd(ans, abs(i - pos[i]))


why are we using GCD HERE ???