0015-3sum.rb

# frozen_string_literal: true

# 15. 3Sum
# https://leetcode.com/problems/3sum
# Medium

=begin
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
=end

# @param {Integer[]} nums
# @return {Integer[][]}
defthree_sum(nums)
nums.sort!
triplets=[]
n=nums.length - 1

(0..n - 2).eachdo |i|
left=i + 1
right=n
target= -nums[i]

nextifi.positive? && nums[i] == nums[i - 1]

whileleft < right
sum=nums[left] + nums[right]
ifsum < target
left += 1
elsifsum > target
right -= 1
elsifsum == target
triplets << [nums[i],nums[left],nums[right]]
left += 1
right -= 1

left += 1whileleft < right && nums[left] == nums[left - 1]
end
end
end

triplets.uniq
end

# **************** #
# TEST #
# **************** #

require"test/unit"
classTest_three_sum < Test::Unit::TestCase
deftest_
assert_equal[[-1, -1,2],[-1,0,1]],three_sum([-1,0,1,2, -1, -4])
assert_equal[],three_sum([0,1,1])
assert_equal[[0,0,0]],three_sum([0,0,0])
end
end