 # XORORED - Editorial

Author: Prasant Kumar
Tester: Aryan Choudhary
Editorialist: Vichitr Gandas

SIMPLE

# PREREQUISITES:

Bitwise Operations, Bit Manipulation

# PROBLEM:

Given an array A of N non-negative integers, you can choose any non-negative integer X and replace every element A_i with (A_i⊕X) Here, denotes the bitwise XOR operation.

Using the above operation exactly once, your goal is to minimize the bitwise OR of the new array. In other words, find X such that (A_1⊕X) ∨ \ldots ∨(A_N⊕X) is minimized, where denotes the bitwise OR operation.

# EXPLANATION

### Observations

• While taking OR, if any of the number has i^{th} bit ON then result will also have i^{th} bit ON.
• If a number has i^{th} bit set, then doing XOR with 2^i resets the i^{th} bit.
• 0⊕ 0 = 0, 0⊕1 = 1, 1⊕0 = 1, 1⊕1=0
• 0∨ 0 = 0, 0∨1 = 1, 1∨0 = 1, 1∨1 =1

### Finding X

Now let’s come back to the problem. We want to do choose an X and do XOR with all array elements. And after this, we calculate OR of all the elements. The task is to minimise the final OR.

We want to find X such that it minimises the final OR. Now let’s look at the problem bitwise. Let’s say we want to find if i^{th} bit in X should be 0 or 1?

There are 3 possible cases:

Case 1: If all the elements of A has i^{th} bit 0.
In this case, X should have i^{th} bit 0.

Why?

If X has i^{th} bit 1 then doing XOR of X with any A_j will turn the i^{th} bit 1. In other words, doing A_j ⊕ X will set the i^{th} bit of A_j as 1. This will further lead to being the i^{th} bit 1 in final OR value which we want to minimise. Hence it should be 0 in X.

Case 2: If all the elements of A has i^{th} bit 1.
In this case, X should have i^{th} bit 1.

Why?

If X has i^{th} bit 1 then doing XOR of X with any A_j will turn the i^{th} bit 0. In other words, doing A_j ⊕ X will reset the i^{th} bit of A_j as 0. And as we are doing XOR of X with every element of A hence it would make it 0 in all the elements. Hence the final OR will have i^{th} bit 0 as all elements have it 0.

Case 3: If some of the elements of A has i^{th} bit 0 and some of them has 1.
In this case, X can have i^{th} bit either 0 or 1. It doesn’t change the final OR value.

Why?

As some of the elements has i^{th} bit 0 and others have 1. If we do not have i^{th} bit 1 in X then the elements which already had it 1 will still have it 1 even after XOR hence final OR value will also have i^{th} bit 1.
Similarly, if we make i^{th} bit 1 in X then after XOR, the elements which earlier had it 1 will have it 0 but the ones which had it 0 will have 1 now. Hence final OR value will also have i^{th} bit 1.

So in both cases, final OR value has i^{th} bit 1 hence it does not matter if i^{th} bit of X is 0 or 1 in this case.

As explained above, just check for every bit of X and find the value of X.

### Alternate Approach to Find X

As we saw in previous approach, i^{th} bit of X remains 0 if its 0 in all the elements. It becomes 1 if its 1 in all the elements. For other cases, we can either have it 0 or 1.
So if we take OR of all the elements then the OR will also have i^{th} bit 0 if it was 0 in all the elements. And 1 otherwise.
Hence that is also a valid X which will give us minimum OR after XOR.
So X = A_1 ∨ A_2 ∨ \ldots ∨ A_N.

Once we have X, do A_i := A_i ⊕ X for all i and finally find OR of all the elements.

# TIME COMPLEXITY:

O(N \cdot \log max(A) ) or O(N) per test case

# SOLUTIONS:

Setter's Solution
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define endl "\n"
int inf=1e9;
signed main(){

ios_base::sync_with_stdio(0) , cin.tie(0);
int t;cin>>t;
assert(t<=5000 and t>=1);
while(t--){
int n;cin>>n;
assert(n<=100 and n>=1);
int arr[n];
int OR=0,AND=-1;
for(int i=0;i<n;i++){
cin>>arr[i];
OR|=arr[i];
AND&=arr[i];
assert(arr[i]<=inf and arr[i]>=0);
}
cout<<AND<<" "<<(OR-AND)<<endl;
}
return 0;
}

Tester'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

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;
}

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--)
{

lli c=0,d=0;
for(auto x:a){
c|=x;
d|=~x;
}
cout<<c<<" "<<(c&d)<<endl;
}   aryanc403();
return 0;
}

Editorialist's Solution
/*
* @author: vichitr
* @date: 25th July 2021
*/

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define fast ios::sync_with_stdio(0); cin.tie(0);

void solve() {
int N; cin >> N;
int A[N];
for (int i = 0; i < N; i++)
cin >> A[i];
int X = 0;
// find X by taking OR of all the elements.
for (int i = 0; i < N; i++) {
X |= A[i];
}
// do XOR and find OR
int ans = 0;
for (int i = 0; i < N; i++) {
ans |= (A[i] ^ X);
}
cout << X << ' ' << ans << '\n';
}

signed main() {
fast;

#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif

int t = 1;
cin >> t;
for (int tt = 1; tt <= t; tt++) {
// cout << "Case #" << tt << ": ";
solve();
}
return 0;
}


If you have other approaches or solutions, let’s discuss in comments.If you have other approaches or solutions, let’s discuss in comments.

I found the max number in the array and took that as X. That solved the problem. I would be grateful if someone explained why that worked.

1 Like

Yeah ,I also select X as the maximum element and got correct answer , Please explain why is it so ?
https://www.codechef.com/viewsolution/49158736

1 Like

Hmmm … can’t find a counter-example, yet 2 Likes

let say bitwise AND of all element A and bitwise OR of all element O.
then for X to be valid A&X should be equal to A and O|X should be equal to O. Every element of the array satisfy this condition.

3 Likes

Did you read the 3 possible cases in finding X as mentioned in the editorial. Only first two are important and they say:

and

And if all elements either has it 0 or 1 then any element can work as X.

5 Likes