3Sum LeetCode Solution

Quiestion:Click Here

Approach:

since we need to find triplet,we fix one nums[i] in given array then we traverse through array for the other two numbers such that their sum is equal to complement of nums[i](i.e,-(nums[i])) using two pointer technique.

we fix low pointer next to the nums[i] and high pointer to the last element.

if nums[i]+nums[low]+nums[high] <0

    increase low pointer

if nums[i]+nums[low]+nums[high] >0

    decrement high pointer

if nums[i]+nums[low]+nums[high] ==0

    add nums[i],nums[low],nums[high] to result list

 

Python Code:

class Solution(object):

    def threeSum(self, nums):

        res = []

        nums.sort()

        for i in range(len(nums)-2):

            if nums[i]>0: break #[7]

            if i>0 and nums[i]==nums[i-1]: continue

            low,high = i+1,len(nums)-1 

            while low < high:

                if nums[i]+nums[low]+nums[high]==0:

                    res.append([nums[i],nums[low],nums[high]])

                    while low<high and nums[low]==nums[low+1]:

                        low+=1

                    while low<high and nums[high]==nums[high-1]:

                        high-=1

                    low+=1

                    high-=1

                elif nums[i]+nums[low]+nums[high]<0:

                    low+=1

                else:

                    high-=1

        return res

click here for code

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