Given an array `nums`

of *n* integers and an integer `target`

, find three integers in `nums`

such that the sum is closest to `target`

. Return the sum of the three integers. You may assume that each input would have exactly one solution.

**Example 1:**

Input:nums = [-1,2,1,-4], target = 1Output:2Explanation:The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

**Constraints:**

`3 <= nums.length <= 10^3`

`-10^3 <= nums[i] <= 10^3`

`-10^4 <= target <= 10^4`

class Solution:
def threeSumClosest(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
nums.sort()
min_ = abs(sum(nums[0:3])-target)
res = nums[0:3]
for i in range(len(nums)-2):
if i > 0 and nums[i] == nums[i-1]:
continue
l, r = i+1, len(nums)-1
while l < r:
s = nums[i] + nums[l] + nums[r]
if s-target<0:
if target-s < min_:
min_ = target-s
res = [nums[i], nums[l], nums[r]]
l += 1
else:
if s-target < min_:
min_ = s-target
res = [nums[i], nums[l], nums[r]]
r -= 1
return sum(res)

The main gist of this question is to reduce the 3 sum problem into a 2 sum problem and using the two-pointer approach to solve it.

Time Complexity: O(n^2)

Space Complexity: O(1)