PROBLEM: (Source)

Given $$$n$$$ $$$d$$$-dimensional vectors $$$v_1, v_2, \cdots, v_n$$$, output $$$i > j$$$ such that $$$v_i \cdot v_j \equiv 0 \mod k$$$, or determine that such indexes do not exist.

Constraints: For 50% cases $$$n \leq 10^5, d \leq 30, 2 \leq k \leq 3$$$, for all others $$$n \leq 10^4, d \leq 100, 2 \leq k \leq 3$$$.

$$$k = 2$$$

The main idea is check for every $$$i$$$ if there exists a $$$j$$$ satisfying the requirement. Keep in mind that here, $$$i$$$ can be treated as a constant. Therefore, requirement is equivalent to negating

The inner summation can be computed quickly using prefix sum. Therefore complexity is O($$$nd$$$) where the $$$n$$$ comes from looping through every $$$i$$$.

$$$k = 3$$$

Now we deal with $$$k = 3$$$. Same idea, but now we don't have the nice property of $$$\not\equiv 0 \implies \equiv 1$$$. However, we can do something clever:

The inner summation can again be stored from previous queries. Therefore the time complexity is $$$O(nd^2)$$$.

Details:

1) The logic above implies that if $$$\sum \not\equiv i-1 \mod 3$$$, there will be a pair of vectors with inner product divisible by $$$3$$$. HOWEVER, it is possible that such pair exists but still bypasses all tests. For example taking $$$d = 1, k = 2, v_1 = 0, v_i = 1 \forall 2\leq i\leq n$$$, the vectors will satisfy that test above. Therefore, it is necessary to shuffle the vectors before applying the check.

2) In the original problem, we also have to restore the index. If we know a given $$$i$$$ fails, we can simply check all $$$j<i$$$ in $$$O(nd)$$$ time.

My implementation can be found here