# GMINEQ-Editorial

Setter: Hriday
Tester: Aryan, Satyam
Editorialist: Devendra Singh

1254

None

# PROBLEM:

You are given an array A of length N containing the elements -1 and 1 only. Determine if it is possible to rearrange the array A in such a way that A_i is not the geometric mean of A_{i-1} and A_{i+1}, for all i such that 2 \leq i \leq N-1.

Y is said to be the geometric mean of X and Z if Y^2 = X \cdot Z.

# EXPLANATION:

An observation: A_i and A_{i+2} cannot be equal as then A_i \cdot A_{i+2} = A_i^2 = 1 = A_{i+1}^2, independent of the value of A_i and A_{i+1} ( as they can be 1 and -1 only )
This means if we fix A_0 all values at even indices can be retrieved using the formula
A_i=-A_{i-2}. Similarly if we fix A_1 all values at odd indices can be retrieved with the same formula.
There are four combinations for first two values and they are [1,1] , [1,-1] , [-1,1] , [-1,-1]. We can generate four different sequences of length N using these four combinations and then check if one of them is an arrangement of the given array.

# TIME COMPLEXITY:

O(N) for each test case.

# SOLUTION:

Tester-1's Solution
/* in the name of Anton */

/*
Compete against Yourself.
Author - Aryan (@aryanc403)
Atcoder library - https://atcoder.github.io/ac-library/production/document_en/
*/

#ifdef ARYANC403
#else
#pragma GCC optimize ("Ofast")
#pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
//#pragma GCC optimize ("-ffloat-store")
#include<bits/stdc++.h>
#define dbg(args...) 42;
#endif

// y_combinator from @neal template https://codeforces.com/contest/1553/submission/123849801
// http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0200r0.html
template<class Fun> class y_combinator_result {
Fun fun_;
public:
template<class T> explicit y_combinator_result(T &&fun): fun_(std::forward<T>(fun)) {}
template<class ...Args> decltype(auto) operator()(Args &&...args) { return fun_(std::ref(*this), std::forward<Args>(args)...); }
};
template<class Fun> decltype(auto) y_combinator(Fun &&fun) { return y_combinator_result<std::decay_t<Fun>>(std::forward<Fun>(fun)); }

using namespace std;
#define fo(i,n)   for(i=0;i<(n);++i)
#define repA(i,j,n)   for(i=(j);i<=(n);++i)
#define repD(i,j,n)   for(i=(j);i>=(n);--i)
#define all(x) begin(x), end(x)
#define sz(x) ((lli)(x).size())
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define endl "\n"

typedef long long int lli;
typedef long double mytype;
typedef pair<lli,lli> ii;
typedef vector<ii> vii;
typedef vector<lli> vi;

const auto start_time = std::chrono::high_resolution_clock::now();
void aryanc403()
{
#ifdef ARYANC403
auto end_time = std::chrono::high_resolution_clock::now();
std::chrono::duration<double> diff = end_time-start_time;
cerr<<"Time Taken : "<<diff.count()<<"\n";
#endif
}

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;
}
assert(l<=x&&x<=r);
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) {
}
long long readIntLn(long long l, long long r) {
}
string readStringLn(int l, int r) {
}
string readStringSp(int l, int r) {
}

assert(getchar()==EOF);
}

vi a(n);
for(int i=0;i<n-1;++i)
return a;
}

bool isBinaryString(const string s){
for(auto x:s){
if('0'<=x&&x<='1')
continue;
return false;
}
return true;
}

// #include<atcoder/dsu>
//     vector<vi> e(n);
//     atcoder::dsu d(n);
//     for(lli i=1;i<n;++i){
//         e[u].pb(v);
//         e[v].pb(u);
//         d.merge(u,v);
//     }
//     assert(d.size(0)==n);
//     return e;
// }

const lli INF = 0xFFFFFFFFFFFFFFFL;

lli seed;
inline lli rnd(lli l=0,lli r=INF)
{return uniform_int_distribution<lli>(l,r)(rng);}

class CMP
{public:
bool operator()(ii a , ii b) //For min priority_queue .
{    return ! ( a.X < b.X || ( a.X==b.X && a.Y <= b.Y ));   }};

void add( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt==m.end())         m.insert({x,cnt});
else                    jt->Y+=cnt;
}

void del( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt->Y<=cnt)            m.erase(jt);
else                      jt->Y-=cnt;
}

bool cmp(const ii &a,const ii &b)
{
return a.X<b.X||(a.X==b.X&&a.Y<b.Y);
}

const lli mod = 1000000007L;
// const lli maxN = 1000000007L;

lli T,n,i,j,k,in,cnt,l,r,u,v,x,y;
lli m;
string s;
vi a;
//priority_queue < ii , vector < ii > , CMP > pq;// min priority_queue .

int main(void) {
ios_base::sync_with_stdio(false);cin.tie(NULL);
// freopen("txt.in", "r", stdin);
// freopen("txt.out", "w", stdout);
// cout<<std::fixed<<std::setprecision(35);
while(T--)
{

for(auto x:a)
assert(abs(x)==1);
sort(all(a));
bool fl=false;
vi b(n);
for(auto p0:{-1,1})
for(auto p1:{-1,1}){
b[0]=p0;
b[1]=p1;
for(lli i=2;i<n;i++)
b[i]=-b[i-2];
sort(all(b));
if(a==b)
fl=true;
}

cout<<(fl?"yEs":"nO")<<endl;
}   aryanc403();
return 0;
}


Tester-2's Solution
#include <bits/stdc++.h>
using namespace std;
#ifndef ONLINE_JUDGE
#define debug(x) cerr<<#x<<" "; _print(x); cerr<<nline;
#else
#define debug(x);
#endif
#define ll long long

/*
------------------------Input Checker----------------------------------
*/

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){
}
}
}

/*
------------------------Main code starts here----------------------------------
*/
ll MAX=100000;
ll tes_sum=0;
vector<string> YS={"YES","yes","yES","YeS","yEs","YEs","Yes","yeS"};
vector<string> NO={"NO","no","No","nO"};
void solve(){
ll x; map<ll,ll> freq;
for(ll i=1;i<n;i++){
freq[x]++;
}
freq[x]++;
assert((freq[-1]+freq[1])==n);
ll l=(n+1)/2,r=n/2;
vector<ll> a={(l+1)/2,l/2},b={(r+1)/2,r/2};
for(auto i:a){
for(auto j:b){
if(i+j==freq[1]){
cout<<YS[rng()%8]<<"\n";
return;
}
}
}
cout<<NO[rng()%4]<<"\n";
return;
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
freopen("error.txt", "w", stderr);
#endif
while(test_cases--){
solve();
}
assert(getchar()==-1);
return 0;
}


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>
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)
{
bool can = false;
int cntone = 0, cntminusone = 0, n, a;
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> a;
cntone += (a == 1);
cntminusone += (a == -1);
}
int tcntone = 0, tcntminusone = 0;
int arr[n];
arr[0] = arr[1] = 1;
tcntone = 2;
for (int i = 2; i < n; i++)
{
arr[i] = -arr[i - 2];
tcntone += (arr[i] == 1);
tcntminusone += (arr[i] == -1);
}
if (tcntone == cntone && tcntminusone == cntminusone)
can = true;
arr[0] = -1, arr[1] = 1;
tcntone = tcntminusone = 1;
for (int i = 2; i < n; i++)
{
arr[i] = -arr[i - 2];
tcntone += (arr[i] == 1);
tcntminusone += (arr[i] == -1);
}
if (tcntone == cntone && tcntminusone == cntminusone)
can = true;
arr[0] = 1, arr[1] = -1;
tcntone = tcntminusone = 1;
for (int i = 2; i < n; i++)
{
arr[i] = -arr[i - 2];
tcntone += (arr[i] == 1);
tcntminusone += (arr[i] == -1);
}
if (tcntone == cntone && tcntminusone == cntminusone)
can = true;
arr[0] = arr[1] = -1;
tcntminusone = 2;
tcntone=0;
for (int i = 2; i < n; i++)
{
arr[i] = -arr[i - 2];
tcntone += (arr[i] == 1);
tcntminusone += (arr[i] == -1);
}
if (tcntone == cntone && tcntminusone == cntminusone)
can = true;
if(can)
cout<<"Yes\n";
else
cout<<"No\n";
return;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
int test = 1;
cin >> test;
while (test--)
sol();
}



Solution based on observation:-

1. For every N>=4 we need 2 different parity elements for every 4 elements in the given sequence
Ex. N =5
(1, -1, -1, 1, 1) 1= 3 -1= 2
(1, 1, -1, -1, 1) 1= 3 -1= 2
(-1, -1, 1, 1, -1) 1= 2 -1= 3
(-1, 1, 1, -1, -1) 1= 2 -1= 3


Here at every 4 elements we require 2 different parity elements (1,1) or (-1,-1)

1. For Every N >=4 if N is divisible by 2 then we can arrange the elements as
Ex. N = 6
6 (1, 1, -1, -1, 1, 1) 1= 4 -1= 2
6 (-1, 1, 1, -1, -1, 1) 1= 3 -1= 3
6 (1, -1, -1, 1, 1, -1) 1= 3 -1= 3
6 (-1, -1, 1, 1, -1, -1) 1= 2 -1= 4

1. For every N >=4 if N %4 == 3 then we need “-1” or “1” to arrange the element in given condition

based on the observation :-
The code :

# cook your dish here
for _ in range(int(input())):
N = int(input())
li = list(map(int, input().split()))
# counting 1
poz = li.count(1)
# counting -1
neg = li.count(-1)
summ = sum(li)
# 1/-1 needed per 4 elements
neededCountPerFour = 2
ans = 0
# handling N =3 separatly
if N == 3:
if summ == 3 or summ == -3:
print('No')
else:
print('Yes')
# logic based on observations
else:
if N%4 == 3:
ans = (neededCountPerFour*(N//4)) +1
else:
ans = (neededCountPerFour*(N//4))
# ans contains the requied no of 1/-1 for the sequence to be become non geometric.
if (ans == poz or ans == neg) or (poz == neg):
print('Yes')
else:
print('No')

1 Like

I tried it with a simpler observation approach.

I observed that:

1. In every consecutive four, there is an equal number of 1 and -1 present.
2. For each case when N is divisible by 4, the difference must be 0.
3. For other cases, the absolute difference between the count of 1 and -1 must be lesser than or equal to 2.
(i) For N%4==1 and N%4==3, the absolute difference must be 1.
(ii) For N%4==2, the absolute difference can be 0 and 2.

Here’s the python code:-

for t in range(int(input())):
n = int(input())
arr = list(map(int,input().split()))
diff = abs(sum(arr))
if diff>2:
print("No")
elif n%4==0 and diff>0:
print("No")
else:
print("Yes")