4Sum - Leetcode 18 - Two Pointers (Python)

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[@GregHogg]

Master Data Structures & Algorithms for FREE at AlgoMap.io/

[@DrFrancis1686]

Awesome explanation, thank you very much!

[@donkeykong5616]

Better than neetcode lol

[@Thebeast_QwQ]

Isn't it kinda misleading to call it a 2 pointer approach when we essentially use n (here 4) pointers (lo and hi are basically pointers too right?)

[@mitranshuraj2811]

Feels kinda bad to understand these topics😂

[@gourav6968]

oi, why is it?

[@mitranshuraj2811]

​@@gourav69682sum, 3sum, 4sum, dp...shall I go on?

[@JoeTan-nq4fq]

If it is a k sum problem, does it mean we need to resort to recursive solution (until we end up with 2 sum problem)?

[@Fen-i3n]

sorry but I dont know how to do 3 sum😮‍💨

[@xingyuxiang1637]

Build a graph and then traverse it. Hashing on nodes, integers, or IDs.
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
n = len(nums)
seen = set()
ans = set()
for i in range(n):
for j in range(i + 1, n):
lastNumber = - nums[i] - nums[j]
if lastNumber in seen:
ans.add((nums[i], nums[j], lastNumber))
seen.add(nums[i])
return ans
class Solution:
def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
nums.sort()
n = len(nums)
seen = set()
ans = set()
for i in range(n):
for j in range(i + 1, n):
for k in range(j + 1, n):
lastNumber = target - nums[i] - nums[j] - nums[k]
if lastNumber in seen:

ans.add((nums[i], nums[j], nums[k], lastNumber))
seen.add(nums[i])
return ans