/**
For all shoes, greedily find the other closest pair near it.
The number of swaps required for that move is the index of the closest pair - current index - number of used shoes in between
The number of used shoes in between can be found using a segtree
*/
#include <bits/stdc++.h>
#include "shoes.h"
#define F0R(i,n) for(auto i = 0; i < (n); i++)
#define FOR(i,a,b) for(auto i = (a); i <= (b); i++)
#define ROF(i,a,b) for(auto i = (a); i >= (b); i--)
#define ll long long
#define pb push_back
#define F first
#define S second
#define vi vector<int>
using namespace std;
const int MAXN = 2e5+5;
int seg[4*MAXN];
bool used[MAXN];
queue<int> leftShoes[MAXN],rightShoes[MAXN];
void update(int node,int ss,int se,int idx) {
if(ss > se || idx > se || idx < ss)return;
if(ss == se) {
++seg[node];
return;
}
int mid = ss+se >> 1;
update(node*2+1,ss,mid,idx);
update(node*2+2,mid+1,se,idx);
seg[node] = seg[2*node+1]+seg[2*node+2];
}
int query(int node,int ss,int se,int qs,int qe) {
if(ss > se || qs > se || qe < ss) return 0;
if(ss >= qs && se <= qe) return seg[node];
int mid = ss+se >> 1;
int q1 = query(node*2+1,ss,mid,qs,qe);
int q2 = query(node*2+2,mid+1,se,qs,qe);
return q1+q2;
}
ll count_swaps(vi a) {
int n = a.size();
F0R(i,n) {
if(a[i] < 0) leftShoes[abs(a[i])].push(i);
else rightShoes[a[i]].push(i);
}
ll ans = 0;
F0R(i,n) {
if(used[i]) continue;
int ni = -1;
if(a[i] < 0) {
ni = rightShoes[abs(a[i])].front();
} else {
ni = leftShoes[abs(a[i])].front();
}
rightShoes[abs(a[i])].pop();
leftShoes[abs(a[i])].pop();
int swapss = ni-i;
swapss -= query(0,0,n-1,i+1,ni-1);
update(0,0,n-1,i);
update(0,0,n-1,ni);
if(a[i] < 0)--swapss;
used[i] = true;
used[ni] = true;
ans += swapss;
}
return ans;
}
