XORR-Editorial

editorial
, ,
1

PROBLEM LINK:

Contest Division 1
Contest Division 2
Contest Division 3
Contest Division 4

Setter: Utkarsh Gupta
Tester: Satyam, Utkarsh Gupta
Editorialist: Devendra Singh

DIFFICULTY:

To be calculated

PREREQUISITES:

Bitwise XOR, C++ Map

PROBLEM:

Chef has two arrays A and B, each of length N.

A pair (i,j) (1 \leq i \lt j \leq N) is said to be a good pair if and only if
A_i \oplus A_j = B_i \oplus B_j

Here, \oplus denotes the bitwise XOR operation.

Determine the number of good pairs.

EXPLANATION:

Let us simplify the equation given in the problem by taking XOR with A_j\oplus B_i on both sides.
A_i \oplus A_j \oplus A_j \oplus B_i = B_i \oplus B_j \oplus A_j \oplus B_i
A_i \oplus B_i = A_j \oplus B_j
\therefore We need to find the number of pair (i,j) (1 \leq i \lt j \leq N) such that A_i \oplus B_i = A_j \oplus B_j
Iterate on the arrays from i=1 to N, keeping track of A_i\oplus B_i with the help of a map data structure. For each index i add the count of A_i\oplus B_i to the answer then update the count of A_i\oplus B_i in the map.

TIME COMPLEXITY:

O(Nlog(N)) or 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];
vector <int> adj[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 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){
    return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
    return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
    return readString(l,r,'\n');
}
string readStringSp(int l,int r){
    return readString(l,r,' ');
}
int maxAi=((1<<30)-1);
int sumN=0;
int A[N],B[N],C[N];
void solve()
{
    int n=readInt(1,300000,'\n');
    sumN+=n;
    assert(sumN<=300000);
    for(int i=1;i<=n;i++)
    {
        if(i==n)
            A[i]=readInt(0,maxAi,'\n');
        else
            A[i]=readInt(0,maxAi,' ');
    }
    for(int i=1;i<=n;i++)
    {
        if(i==n)
            B[i]=readInt(0,maxAi,'\n');
        else
            B[i]=readInt(0,maxAi,' ');
    }
    set <int> s;
    map <int,ll> cnt;
    for(int i=1;i<=n;i++)
    {
        C[i]=(A[i]^B[i]);
        s.insert(C[i]);
        cnt[C[i]]++;
    }
    ll ans=0;
    for(auto it:s)
        ans+=((cnt[it]*(cnt[it]-1))/2);
    cout<<ans<<'\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);
    int T=readInt(1,100000,'\n');
    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); }
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void sol(void)
{
    ll n, ans = 0;
    cin >> n;
    map<ll, ll> cnt;
    vll a(n), b(n);
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
    }
    for (int i = 0; i < n; i++)
    {
        cin >> b[i];
        ans += cnt[a[i] ^ b[i]];
        cnt[b[i] ^ a[i]]++;
    }
    cout << ans << '\n';
    return;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL), cout.tie(NULL);
    int test = 1;
    cin >> test;
    while (test--)
        sol();
}
3 Likes
2 
#include <bits/stdc++.h>
#define ull unsigned long long int
#define ui unsigned int
#define ll long long int
#define f(i, n) for (ll i = 0; i < n; i++)
const ll m = 10e9 + 7;
// const ll N = 10e5 + 5;
using namespace std;

ll binpow(ll a, ll b, ll m)
{
    a %= m;
    ll res = 1;
    while (b > 0)
    {
        if (b & 1)
            res = res * a % m;
        a = a * a % m;
        b >>= 1;
    }
    return res;
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    ll t;
    cin >> t;
    while (t--)
    {
        ll n;
        cin >> n;
        ll arr1[n];
        ll arr2[n];
        f(i, n) cin >> arr1[i];
        f(i, n) cin >> arr2[i];
        unordered_map<ll, ll> map;
        for (int i = 0; i < n; i++)
            map[arr1[i] ^ arr2[i]]++;
        int ans = 0;
        for(auto& i : map){
            ans += ((i.second*(i.second-1))/2);
        }
        cout << ans << '\n';
    }
    return 0;
}
please anyone tell why this is giving WA in one testcase.
3

Change int ans to long long ans as N is large and answer could be very large ((N*(N-1))//2) exceeding the range of int

2 Likes
4

We have to check Ai ^ Bi == Aj ^ Bj then why we are considering only Ai ^ Bi

5

// whats wrong in my code ?

#include

using namespace std;

int main()

{

int t;

cin >> t;

while (t--)

{

    int counter = 0;

    int n;

    cin >> n;

    int a[n], b[n];

    for (int i = 0; i < n; i++)

        cin >> a[i];

    for (int i = 0; i < n; i++)

        cin >> b[i];

    for (int i = 0; i < n - 1; i++)

    {

        for (int j = i + 1; j < n; j++)

        {

            if ((a[i] ^ a[j + 1]) == (b[i] ^ b[j + 1]))

            {

                counter++;

            }

        }

    }

    cout << counter << "\n";

}

}

6

the time complexity is O(nĖ†2) which would end up in TLE.
Also if you are looping j from i+1 to n, then indexing j+1 will result in a segmentation fault as well.

7

I was using the standard bruteforce approach lol.