mirror of
https://github.com/TheAlgorithms/Ruby
synced 2024-12-26 21:58:56 +01:00
88 lines
2.2 KiB
Ruby
88 lines
2.2 KiB
Ruby
# Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] ..
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# .. such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
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# Notice that the solution set must not contain duplicate triplets.
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# Example 1:
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# Input: nums = [-1,0,1,2,-1,-4]
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# Output: [[-1,-1,2],[-1,0,1]]
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# Example 2:
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# Input: nums = []
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# Output: []
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# Example 3:
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# Input: nums = [0]
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# Output: []
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# Constraints:
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# 0 <= nums.length <= 3000
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#-105 <= nums[i] <= 105
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# Two Pointer Approach - O(n) Time / O(1) Space
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# Return edge cases.
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# Sort nums, and init ans array
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# For each |val, index| in nums:
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# if the current value is the same as last, then go to next iteration
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# init left and right pointers for two pointer search of the two sum in remaining elements of array
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# while left < right:
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# find current sum
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# if sum > 0, right -= 1
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# if sum < 0, left += 1
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# if it's 0, then add the values to the answer array, and set the left pointer to the next valid value ..
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# .. (left += 1 while nums[left] == nums[left - 1] && left < right)
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# Return ans[]
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# @param {Integer[]} nums
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# @return {Integer[][]}
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def three_sum(nums)
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# return if length too short
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return [] if nums.length < 3
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# sort nums, init ans array
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nums = nums.sort
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ans = []
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# loop through nums
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nums.each_with_index do |val, ind|
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# if the previous value is the same as current, then skip this iteration as it would create duplicates
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next if ind > 0 && nums[ind] == nums[ind - 1]
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# init & run two pointer search
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left = ind + 1
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right = nums.length - 1
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while left < right
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# find current sum
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sum = val + nums[left] + nums[right]
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# decrease sum if it's too great, increase sum if it's too low
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if sum > 0
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right -= 1
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elsif sum < 0
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left += 1
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# if it's zero, then add the answer to array and set left pointer to next valid value
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else
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ans << [val, nums[left], nums[right]]
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left += 1
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left += 1 while nums[left] == nums[left - 1] && left < right
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end
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end
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end
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# return answer
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ans
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end
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nums = [-1, 0, 1, 2, -1, -4]
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print three_sum(nums)
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# Output: [[-1,-1,2],[-1,0,1]]
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nums = []
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print three_sum(nums)
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# Output: []
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nums = [0]
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print three_sum(nums)
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# Output: []
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