Problem description:
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:

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Given array nums = [-1, 2, 1, -4], and target = 1.

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

Solution:

Sort the vector and then no need to run O(N^3) algorithm as each index has a direction to move.

The code starts from this formation.

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----------------------------------------------------
^  ^                                               ^
|  |                                               |
|  +- second                                     third
+-first

if nums[first] + nums[second] + nums[third] is smaller than the target, we know we have to increase the sum. so only choice is moving the second index forward.

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----------------------------------------------------
^    ^                                             ^
|    |                                             |
|    +- second                                   third
+-first

if the sum is bigger than the target, we know that we need to reduce the sum. so only choice is moving ‘third’ to backward. of course if the sum equals to target, we can immediately return the sum.

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----------------------------------------------------
^    ^                                          ^
|    |                                          |
|    +- second                                third
+-first

when second and third cross, the round is done so start next round by moving ‘first’ and resetting second and third.

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----------------------------------------------------
  ^    ^                                           ^
  |    |                                           |
  |    +- second                                 third
  +-first

while doing this, collect the closest sum of each stage by calculating and comparing delta. Compare abs(target-newSum) and abs(target-closest). At the end of the process the three indexes will eventually be gathered at the end of the array.

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----------------------------------------------------
                                         ^    ^    ^
                                         |    |    `- third
                                         |    +- second
                                         +-first

if no exactly matching sum has been found so far, the value in closest will be the answer.

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class Solution:
    def threeSumClosest(self, nums: List[int], target: int) -> int:
        nums.sort()
        res = nums[0] + nums[1] + nums[2]
        
        for i in range(len(nums)-2):
            l, r = i+1, len(nums)-1
            while l < r:
                curSum = nums[i] + nums[l] + nums[r]
                if curSum == target: return curSum
                if abs(curSum - target) < abs(res-target):
                    res = curSum
                if curSum < target:
                    l += 1
                else:
                    r -= 1
        return res
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class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        
        if(nums.size() < 3) return 0;
        int closest = nums[0]+nums[1]+nums[2]; //pick a sum value to compare
        sort(nums.begin(), nums.end()); //sort the array, takes O(log n)
        for(int first = 0 ; first < nums.size()-2 ; ++first) {
            if(first > 0 && nums[first] == nums[first-1]) continue; //skip the duplicate items
            int second = first+1; //2nd pointer go toward left
            int third = nums.size()-1; //3rdd pointer go toward right         
            while(second < third) { //use 1st pointer as index, find the set has closest sum
                int curSum = nums[first]+nums[second]+nums[third];
                if(curSum == target) return curSum;
                if(abs(target-curSum)<abs(target-closest)) {
                    closest = curSum;
                }
                if(curSum > target) { //because the array is sorted, we can decide where the pointer should move.
                    --third;
                } else {
                    ++second;
                }
            }
        }
        return closest;
    }
};