Submission #2704246

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Source Code Expand

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 200001;

string s;
deque<int> oc[26];
bool done[MX];
int good[MX];

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> s;
    F0R(i,sz(s)) oc[s[i]-'a'].pb(i);
    
    int z = 0;
    F0R(i,26) if (sz(oc[i])&1) z ++;
    if (z > 1) {
        cout << -1;
        exit(0);
    }
    int nex = 0;
    F0R(i,sz(s)) if (!done[i]) {
        if (sz(oc[s[i]-'a']) == 1) {
            good[i] = sz(s)/2;
        } else {
            good[i] = nex;
            good[oc[s[i]-'a'].back()] = sz(s)-1-nex;
            nex ++;
            
            done[oc[s[i]-'a'].back()] = 1;
            oc[s[i]-'a'].pop_front();
            oc[s[i]-'a'].pop_back();
        }
    }
    
    Tree<int> T;
    
    ll ans = 0;
    F0R(i,sz(s)) {
        ans += sz(T)-T.order_of_key(good[i]);
        T.insert(good[i]);
    }
    cout << ans;
}

// read the question correctly (is y a vowel? what are the exact constraints?)
// look out for SPECIAL CASES (n=1?) and overflow (ll vs int?) ARRAY OUT OF BOUNDSS

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