Two pointers technique

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Two pointers is typically used for searching pairs in a sorted array satisfying some condition in linear time.

[Leetcode 15] 3Sum

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

Example 1:

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Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation: 
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.

Example 2:

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Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.

Example 3:

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Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.

Constraints:

  • 3 <= nums.length <= 3000
  • -10^5 <= nums[i] <= 10^5

Solution:

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class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        triplets = set() # automatically remove duplicates

        for i in range(len(nums) - 2):
            j = i + 1
            k = len(nums) - 1
            while j < k:
                sum = nums[i] + nums[j] + nums[k]
                if sum > 0:
                    k -= 1
                elif sum < 0:
                    j += 1
                else:
                    triplets.add((nums[i],nums[j],nums[k]))
                    j += 1
                    k -= 1

        return triplets

Reference