The naive algorithm is $O(N^4^)$ try all the possible combinations of four numbers.
A less naive algorithm brute forces through all $n^3^$ combinations of three numbers and then efficiently checks if their sum is in the original array using a hash table. This algorithm has $O(n^3^)$ complexity.

By now, you’ve probably started thinking how meet in the middle can be applied to this problem. The trick is to rewrite the equation as $a + b = -(c + d)$. Store all the $n^2^$ sums $a + b$ in a hash set $S$. Then iterate through all $n^2^$ combinations for c and d and check if $S$ contains $-(c + d)$.

By now, you’ve probably started thinking how meet in the middle can be applied to this problem. The trick is to rewrite the equation as $a + b = -(c + d)$. Store all the $n^2^$ sums $a + b$ in a hash set $S$. Then iterate through all $n^2^$ combinations for c and d and check if $S$ contains $ -(c + d)$.

Here's how the code looks
== code(c) |