[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:

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:

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

Example 3:

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

Constraints:

• 3 <= nums.length <= 3000
• -105 <= nums[i] <= 105

Solution:

def three_sum(nums)
    nums.sort!
    result = []

    (0..nums.length - 2).each do |i|
        next if i > 0 && nums[i] == nums[i - 1]

        left = i + 1
        right = nums.length - 1

        while left < right
            sum = nums[i] + nums[left] + nums[right]

            if sum == 0
                result << [nums[i], nums[left], nums[right]]

                left += 1 while left < right && nums[left] == nums[left + 1]
                right -= 1 while left < right && nums[right] == nums[right - 1]

                left += 1
                right -= 1
            elsif sum > 0
                right -= 1
            else
                left += 1
            end
        end
    end

    result
end