ENDSORTED-Editorial

PROBLEM LINK:

Contest Division 1
Contest Division 2
Contest Division 3
Contest Division 4

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

DIFFICULTY:

1049

PREREQUISITES:

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];
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(2,100000,'\n');
    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)
            P[i]=readInt(1,n,' ');
        else
            P[i]=readInt(1,n,'\n');
        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);
    int T=readInt(1,1000,'\n');
    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); }
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
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();
}