[LeetCode] 16. 3Sum Closest

Problem : https://leetcode.com/problems/3sum-closest/

Use the same two pointers approach as the 3Sum quiz.

Time Complexity :  O ( N * LogN + N ** 2 )

Space Complexity: O ( log N ) or O ( N ), depending on the sort algorithm. 

class Solution(object):
    def threeSumClosest(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        
        nums.sort()
        N = len(nums)
        
        result = nums[0] + nums[1] + nums[2] # assume the init value of result
        
        for i in range(N-2):
            if i > 0 and nums[i] == nums[i-1]:
                continue
                
            left, right = i + 1, N-1
                        
            while left < right:
                tmp = sum([nums[i], nums[left], nums[right]])
                
                if abs(result - target) > abs(tmp - target):
                    result = tmp
                
                if tmp == target:
                    return target
                
                if tmp < target:
                    j = left + 1
                    while j < right and nums[j] == nums[left]:
                        j += 1
                    left = j
                else:
                    k = right - 1
                    while k > left and nums[k] == nums[right]:
                        k -= 1
                    right = k
                    
        return result

Edited of 07/27/2021. Update result variable's init value.