The editorial says:

"There is very simple solution. Let's check that for all i a[i] = b[i]. If this condition is true we should print "Poor Alex"."

You might be confusing this statement with an "if and only if". The implication in this statement is only for one side. It doesn't say that the only case where we should print "Poor Alex" is if $$$a_i$$$ $$$=$$$ $$$b_i$$$ $$$\forall i$$$. The rest of the editorial gives the full condition that should be satisfied for the answer to be "Poor Alex".

Neel_Knight you are right.

My Solution :

void solve()
{
    int n;
    cin >> n;

    vpi v(n);

    for (int i = 0; i < n; i++)
    {
        int x, y;
        cin >> x >> y;

        v[i] = {x, y};
    }

    sort(v.begin(), v.end(), myComp);
    int pos;
    for (int i = 0; i < n - 1; i++)
    {
        if (v[i].first != v[i + 1].first && v[i].second < v[i + 1].second)
        {
            cout << "Happy Alex";
            return;
        }
    }

    cout << "Poor Alex";
    return;
}

International Master's solution:

void solve()
{
 int n;
    scanf("%d", &n);
    int a, b;
    for (int i = 0; i < n; i++)
    {
        scanf("%d %d", &a, &b);
        if (a != b)
        {
            printf("Happy Alex");
            return ;
        }
    }
    printf("Poor Alex");
return;
}

Both print different outputs in this case

2
2 1
3 2

Mine prints "Poor Alex" and his code prints "Happy Alex"

Mine and his submission both are accepted.

His code just checks if there exists any pair for which ai != bi and prints "Happy Alex". What's going on??