Problem
Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.
To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of 2**28 to 2**28  1 and the result is guaranteed to be at most 2**31  1.
Example:
Input: A = [ 1, 2] B = [2,1] C = [1, 2] D = [ 0, 2]
Output: 2

Explanation:
The two tuples are:
 (0, 0, 0, 1) > A[0] + B[0] + C[0] + D[1] = 1 + (2) + (1) + 2 = 0
 (1, 1, 0, 0) > A[1] + B[1] + C[0] + D[0] = 2 + (1) + (1) + 0 = 0
Thought Process
 brutal force won't work as 500**4 or 500**3 will exhaust most computers
 this will be an example of compromising space for time
 create a hash table of all possible sums of C and D (Space complexity of O(n**2)) * need a ht of n**2 * k: sum results, v: appearances, time complexity is O(n**2)
 For sum of A, B: * find if sum exists in ht, if it exists, return ht[sum] * else return 0
White Board
Below is the white board:
Code
class Solution:
def _create_ht(self, num, ht):
assert(type(ht)) is dict
if num in ht:
ht[num] += 1
else:
ht[num] = 1
return ht
def four_sum(self, A, B, C, D):
ht = {}
for i in range(len(C)):
for j in range(len(D)):
self._create_ht(C[i] + D[j], ht)
res = 0
for i in range(len(A)):
for j in range(len(B)):
if  A[i]  B[j] in ht:
res += ht[ A[i]  B[j]]
return res
def test():
A = [ 1, 2]
B = [2,1]
C = [1, 2]
D = [ 0, 2]
fs = Solution().four_sum(A,B,C,D)
print(fs)
test()
2
Complexity
 Time complexity is O(n**2) since I used a 2level nested loop
 Space complexity is O(n**2) since I used a ht based off two indices
Optimization
I have yet to figure out a more optimal solution at this moment.