A slightly improved algorithm brute forces through all $n^3^$ three number combinations and efficiently checks if -(a + b + c) is in the original array using a hash table. This algorithm has $O(n^3^)$ complexity.

By now, you’ve probably started wondering how the meet in the middle technique could be applied here. The insight comes from rewriting $a + b + c + d = 0$ as $a + b = -(c + d)$.

By now, you’re probably wondering how the meet in the middle technique can be applied here. The critical insight comes from rewriting $a + b + c + d = 0$ as $a + b = -(c + d)$.

Now we store all $n^2^$ 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