# ENDSORTED-Editorial

Setter:Utkarsh Gupta
Tester: Manan Grover, Abhinav Sharma
Editorialist: Devendra Singh

1049

None

# PROBLEM:

Chef considers a permutation P of \{1, 2, 3, \dots, N\} End Sorted if and only if P_1 = 1 and P_N = N.

Chef is given a permutation P.

In one operation Chef can choose any index i \ (1 \leq i \leq N-1) and swap P_i and P_{i+1}. Determine the minimum number of operations required by Chef to make the permutation P End Sorted.

Note: An array P is said to be a permutation of \{1, 2, 3, \dots, N\} if P contains each element of \{1, 2, 3, \dots, N\} exactly once.

# EXPLANATION:

Let id_1 be the position of 1 in the permutation and id_N be the position of N in the permutation. We want 1 at position 1 and N at position N in the permutation. If id_1 is smaller than id_N we will have to make at least id_1-1 swaps to get 1 at position 1 and N-id_N swaps to get N at position N in the permutation. But if id_1>id_n we will have to do id_1-1 swaps to get 1 at position 1 in the permutation. One of these swaps, swaps N and 1 hence id_N is increased by 1. Therefore the answer in this case is id_1-1+N-(id_N+1).

# TIME COMPLEXITY:

O(N) for each test case.

# SOLUTION:

Setter's solution
//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];
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){
}
}
}
int sumN=0;
void solve()
{
sumN+=n;
assert(sumN<=300000);
int P[n+1]={0};
int indx[n+1]={0};
int mark[n+1]={0};
for(int i=1;i<=n;i++)
{
if(i!=n)
else
indx[P[i]]=i;
mark[P[i]]=1;
}
for(int i=1;i<=n;i++)
assert(mark[i]==1);
int a=indx[1];
int b=indx[n];
int ans=(a-1)+(n-b);
if(a>b)
ans--;
cout<<ans<<'\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);
while(T--)
solve();
assert(getchar()==-1);
cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}

Editorialist's Solution
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define pb push_back
#define all(_obj) _obj.begin(), _obj.end()
#define F first
#define S second
#define pll pair<ll, ll>
#define vll vector<ll>
ll INF = 1e18;
const int N = 1e5 + 11, mod = 1e9 + 7;
ll max(ll a, ll b) { return ((a > b) ? a : b); }
ll min(ll a, ll b) { return ((a > b) ? b : a); }
void sol(void)
{
int n, id1, idn;
cin >> n;
vll p(n);
for (int i = 0; i < n; i++)
{
cin >> p[i];
if (p[i] == 1)
id1 = i;
else if (p[i] == n)
idn = i;
}
if (id1 < idn)
cout << id1 + n - 1 - idn << '\n';
else
cout << id1 + n - 1 - idn - 1 << '\n';

return;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
int test = 1;
cin >> test;
while (test--)
sol();
}



how it would be