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Two Pointers

3-Sum

medium

DESCRIPTION (credit Leetcode.com)

Given an input integer array nums, write a function to find all unique triplets [nums[i], nums[j], nums[k]] such that i, j, and k are distinct indices, and the sum of nums[i], nums[j], and nums[k] equals zero. Ensure that the resulting list does not contain any duplicate triplets.

Input:


nums = [-1,0,1,2,-1,-1]

Output:


[[-1,-1,2],[-1,0,1]]

Explanation: Both nums[0], nums[1], nums[2] and nums[1], nums[2], nums[4] both include [-1, 0, 1] and sum to 0. nums[0], nums[3], nums[4] ([-1,-1,2]) also sum to 0.

Since we are looking for unique triplets, we can ignore the duplicate [-1, 0, 1] triplet and return [[-1, -1, 2], [-1, 0, 1]].

The order of the triplets and the order of the elements within the triplets do not matter.

Solution


class Solution:
    def threeSum(self, nums: List[int]):
      nums.sort()
      result = []
      for i in range(len(nums) - 2):
        if i > 0 and nums[i] == nums[i - 1]:
          continue
        left = i + 1
        right = len(nums) - 1
        while left < right:
          total = nums[i] + nums[left] + nums[right]
          if total < 0:
            left += 1
          elif total > 0:
            right -= 1
          else:
            result.append([nums[i], nums[left], nums[right]])
            while left < right and nums[left] == nums[left + 1]:
              left += 1
            while left < right and nums[right] == nums[right - 1]:
              right -= 1
            left += 1
            right -= 1
      return result

Result

3 sum

0 / 18

Explanation

We can leverage the two-pointer technique to solve this problem by first sorting the array. We can then iterate through each element in the array. The problem then reduces to finding two numbers in the rest of the array that sum to the negative of the current element, which follows the same logic as the Two Sum (Sorted Array) problem from the overview.

Result

Since our first triplet sums to 0, we can add it to our result set.


class Solution:
    def threeSum(self, nums: List[int]):
      nums.sort()
      result = []
      for i in range(len(nums) - 2):
        if i > 0 and nums[i] == nums[i - 1]:
          continue
        left = i + 1
        right = len(nums) - 1
        while left < right:
          total = nums[i] + nums[left] + nums[right]
          if total < 0:
            left += 1
          elif total > 0:
            right -= 1
          else:
            result.append([nums[i], nums[left], nums[right]])
            while left < right and nums[left] == nums[left + 1]:
              left += 1
            while left < right and nums[right] == nums[right - 1]:
              right -= 1
            left += 1
            right -= 1
      return result

Result

3 sum

0 / 5

Avoiding Duplicates

As soon as we find a triplet that sums to 0, we can add it to our result set. We then have to move our left and right pointers to look for the next triplet while avoiding duplicate triplets. We can do this by moving the left and right pointers until they point to different numbers than the ones they were pointing to before.

Here we move the left pointer once until it reaches the last -1 in the array. Then, we can move both the left and right pointers so that they both point to new numbers.


class Solution:
    def threeSum(self, nums: List[int]):
      nums.sort()
      result = []
      for i in range(len(nums) - 2):
        if i > 0 and nums[i] == nums[i - 1]:
          continue
        left = i + 1
        right = len(nums) - 1
        while left < right:
          total = nums[i] + nums[left] + nums[right]
          if total < 0:
            left += 1
          elif total > 0:
            right -= 1
          else:
            result.append([nums[i], nums[left], nums[right]])
            while left < right and nums[left] == nums[left + 1]:
              left += 1
            while left < right and nums[right] == nums[right - 1]:
              right -= 1
            left += 1
            right -= 1
      return result

Result

[-1, -1, 2]

add triplet to output

0 / 2

Here we can do another iteration of the Two Sum problem using the new positions of the left and right pointers.


class Solution:
    def threeSum(self, nums: List[int]):
      nums.sort()
      result = []
      for i in range(len(nums) - 2):
        if i > 0 and nums[i] == nums[i - 1]:
          continue
        left = i + 1
        right = len(nums) - 1
        while left < right:
          total = nums[i] + nums[left] + nums[right]
          if total < 0:
            left += 1
          elif total > 0:
            right -= 1
          else:
            result.append([nums[i], nums[left], nums[right]])
            while left < right and nums[left] == nums[left + 1]:
              left += 1
            while left < right and nums[right] == nums[right - 1]:
              right -= 1
            left += 1
            right -= 1
      return result

Result

[-1, -1, 2]

move both pointers

0 / 3

At this point our left and right pointers have crossed, so we can move our iterator to the next number in the array.

Avoiding Duplicates II

In this case, since the next number in the array is the same as the previous number, we can skip it. We can do this by moving our iterator until it points to a new number.


class Solution:
    def threeSum(self, nums: List[int]):
      nums.sort()
      result = []
      for i in range(len(nums) - 2):
        if i > 0 and nums[i] == nums[i - 1]:
          continue
        left = i + 1
        right = len(nums) - 1
        while left < right:
          total = nums[i] + nums[left] + nums[right]
          if total < 0:
            left += 1
          elif total > 0:
            right -= 1
          else:
            result.append([nums[i], nums[left], nums[right]])
            while left < right and nums[left] == nums[left + 1]:
              left += 1
            while left < right and nums[right] == nums[right - 1]:
              right -= 1
            left += 1
            right -= 1
      return result

Result

[-1, -1, 2]

[-1, 0, 1]

move both pointers

0 / 3

And we're ready to start the Two Sum algorithm again, so we reset our left and right pointers, and start the algorithm.


class Solution:
    def threeSum(self, nums: List[int]):
      nums.sort()
      result = []
      for i in range(len(nums) - 2):
        if i > 0 and nums[i] == nums[i - 1]:
          continue
        left = i + 1
        right = len(nums) - 1
        while left < right:
          total = nums[i] + nums[left] + nums[right]
          if total < 0:
            left += 1
          elif total > 0:
            right -= 1
          else:
            result.append([nums[i], nums[left], nums[right]])
            while left < right and nums[left] == nums[left + 1]:
              left += 1
            while left < right and nums[right] == nums[right - 1]:
              right -= 1
            left += 1
            right -= 1
      return result

Result

[-1, -1, 2]

[-1, 0, 1]

initialize pointers

0 / 2

Termination

Our algorithm terminates when i reaches the 2nd to last element in the array. This is because we need at least 3 elements to form a triplet.


class Solution:
    def threeSum(self, nums: List[int]):
      nums.sort()
      result = []
      for i in range(len(nums) - 2):
        if i > 0 and nums[i] == nums[i - 1]:
          continue
        left = i + 1
        right = len(nums) - 1
        while left < right:
          total = nums[i] + nums[left] + nums[right]
          if total < 0:
            left += 1
          elif total > 0:
            right -= 1
          else:
            result.append([nums[i], nums[left], nums[right]])
            while left < right and nums[left] == nums[left + 1]:
              left += 1
            while left < right and nums[right] == nums[right - 1]:
              right -= 1
            left += 1
            right -= 1
      return result

Result

[-1, -1, 2]

[-1, 0, 1]

move right pointer backward

0 / 2

Test Your Knowledge

Complexity Analysis

Time Complexity: O(n²) where n is the length of the input array. This is due to the nested loops in the algorithm. We perform n iterations of the outer loop, and each iteration takes O(n) time to use the two-pointer technique.

Space Complexity: O(n²) where n is the length of the input array. We need to store all distinct triplets that sum to 0, which can be at most O(n²) triplets.

Next: Triangle Numbers

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