This problem is very similiar to sliding window problems. For each number a, we could have 2 pointer i and j. i starts from 0 and j starts from n - 1. And do a search to get the closest value to 0.
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
ans = set()
for i in range(len(nums) - 2):
j = i + 1
k = len(nums) - 1
while k > j:
s = nums[i] + nums[j] + nums[k]
if s == 0:
ans.add((nums[i], nums[j], nums[k]))
j += 1
elif s < 0:
j += 1
else:
k -= 1
return list(ans)