# MINFLIPS-Editorial

Setter: Utkarsh Gupta
Tester: Nishank Suresh, Abhinav sharma
Editorialist: Devendra Singh

781

None

# PROBLEM:

Chef has an array of length N consisting of 1 and −1 only.

In one operation Chef can choose any index and multiply it by −1.

What is the minimum number of operations required to make sum of the array equal to 0. Output −1 if the sum of the array cannot be made 0.

# EXPLANATION:

In the final array the count of -1 and 1 has to be equal to make the sum equal to 0. Hence if N is odd, it is not possible to make the sum 0. Otherwise take the count of 1 and -1 and whichever is greater, decrease its count to \frac{N}{2} by using the operation defined above.

# 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){
}
}
}
void solve()
{
int A[N+1]={0};
int sum=0;
for(int i=1;i<=N;i++)
{
if(i==N)
else
assert(A[i]!=0);
sum+=A[i];
}
if(N%2==1)
{
cout<<-1<<'\n';
return;
}
sum=abs(sum);
cout<<(sum/2)<<'\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;
cin >> n;
int cnt1 = 0;
vll v(n);
for (int i = 0; i < n; i++)
{
cin >> v[i];
cnt1 += (v[i] == 1);
}
if (n & 1)
cout << -1 << '\n';
else if (cnt1 >= n / 2)
cout << cnt1 - n / 2 << '\n';
else
cout << n / 2 - cnt1 << '\n';

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