Getting WA, created own test cases too but not getting approved answers - c++

Question

Link of Question : https://www.codechef.com/JULY20B/problems/PTMSSNG
Question Statement
Chef has N axis-parallel rectangles in a 2D Cartesian coordinate system. These rectangles may intersect, but it is guaranteed that all their 4N vertices are pairwise distinct.
Unfortunately, Chef lost one vertex, and up until now, none of his fixes have worked (although putting an image of a point on a milk carton might not have been the greatest idea after all…). Therefore, he gave you the task of finding it! You are given the remaining 4N−1 points and you should find the missing one.
Can anyone suggest where I'm going wrong or update my code or share a few test cases.
#include <iostream>
#include <vector>
#include <algorithm>
#include <utility>
#define ll long long
using namespace std;
int main()
{
int t;
cin >> t;
for (int i = 0; i < t; i++)
{
vector<pair<ll, ll>> v;
ll n, m, a;
bool checkx = false;
cin >> n;
m = 4 * n - 1;
ll x[m], y[m];
ll c, d;
a = (m - 1) / 2;
for (ll i = 0; i < m; i++)
{
cin >> x[i] >> y[i];
v.push_back(make_pair(x[i], y[i]));
}
sort(v.begin(), v.end());
for (ll i = a; i >= 1; --i)
{
if (v[2 * i].first != v[2 * i - 1].first)
{
c = v[2 * i].first;
checkx = true;
if ((2 * i) % 4 == 0 && i >= 2)
{
if (v[2 * i].second == v[2 * i + 1].second)
{
d = v[2 * i + 2].second;
}
else
{
d = v[2 * i + 1].second;
}
}
else
{
if (v[2 * i].second != v[2 * i - 1].second)
{
d = v[2 * i - 1].second;
}
else
{
d = v[2 * i - 2].second;
}
}
break;
}
}
if (checkx)
{
cout << c << " " << d;
}
else
{
if (v[0].second == v[1].second)
{
d = v[2].second;
}
else
{
d = v[1].second;
}
cout << v[0].first << " " << d;
}
cout << endl;
}
return 0;
}

Answer 1

You don't need to do such complex things. Just input your x and y vectors and xor every element of each vector. The final value will be the required answer.
LOGIC :
(a,b)------------------(c,b)
| |
| |
| |
| |
(a,d)------------------(c,d)
See by this figure, each variable (a, b, c, d) occurs even number of times. This "even thing" will also be true for the N rectangles. Hence, you have to find the values of x and y which are occurring odd number of times.
To find the odd one out in such cases, the best trick is to xor every element of the vector. This works because of these properties of xor : k xor k = 0 and k xor 0 = k.
CODE:
#include <functional>
#include <iostream>
#include <numeric>
#include <vector>
signed main() {
std::size_t t, n;
std::cin >> t;
while (t--) {
std::cin >> n;
n = 4 * n - 1;
std::vector<int> x(n), y(n);
for (std::size_t i = 0; i < n; ++i)
std::cin >> x.at(i) >> y.at(i);
std::cout << std::accumulate(x.begin(), x.end(), 0L, std::bit_xor<int>()) << ' '
<< std::accumulate(y.begin(), y.end(), 0L, std::bit_xor<int>()) << '\n';
}
return 0;
}

Answer 2

here is a test case that your code doesn't work:
1
2
1 1
1 4
4 6
6 1
9 6
9 3
4 3
the output of your code is (6,3),but it should be (6,4).
I guess you can check more cases where the rectangles intersects.

Answer 3

from functools import reduce
for _ in range(int(input())):
n=int(input())
li=[]
li1=[]
for i in range(4*n-1):
m,n=map(int,input().split())
li.append(m)
li1.append(n)
r =reduce(lambda x, y: x ^ y,li)
print(r,end=' ')
r =reduce(lambda x, y: x ^ y,li1)
print(r,end=' ')
print()