Miroirs/TheAlgorithms-Ruby
1
0
Fork 0
mirror of https://github.com/TheAlgorithms/Ruby synced 2025-01-14 08:01:05 +01:00
Code Issues Projects Releases Packages Wiki Activity

add formatting and example outputs for quick debugging

Browse source
This commit is contained in:
Vitor Oliveira 2021-09-03 12:33:56 -07:00 committed by GitHub
parent bb967fba48
commit bd92198fc8
No known key found for this signature in database
GPG key ID: 4AEE18F83AFDEB23
1 changed files with 76 additions and 68 deletions
Show stats Download patch file Download diff file Expand all files Collapse all files

126
data_structures/arrays/3sum.rb
View file

@ -1,80 +1,88 @@
#Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] .. # 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. # .. 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. # Notice that the solution set must not contain duplicate triplets.
#Example 1: # Example 1:
#Input: nums = [-1,0,1,2,-1,-4] # Input: nums = [-1,0,1,2,-1,-4]
#Output: [[-1,-1,2],[-1,0,1]] # Output: [[-1,-1,2],[-1,0,1]]
#Example 2: # Example 2:
#Input: nums = [] # Input: nums = []
#Output: [] # Output: []
#Example 3: # Example 3:
#Input: nums = [0] # Input: nums = [0]
#Output: [] # Output: []
#Constraints: # Constraints:
#0 <= nums.length <= 3000 # 0 <= nums.length <= 3000
#-105 <= nums[i] <= 105 #-105 <= nums[i] <= 105
# Two Pointer Approach - O(n) Time / O(1) Space
# Return edge cases.
# Sort nums, and init ans array
#Two Pointer Approach - O(n) Time / O(1) Space # For each |val, index| in nums:
#Return edge cases. # if the current value is the same as last, then go to next iteration
#Sort nums, and init ans array # init left and right pointers for two pointer search of the two sum in remaining elements of array
#For each |val, index| in nums: # while left < right:
#if the current value is the same as last, then go to next iteration # find current sum
#init left and right pointers for two pointer search of the two sum in remaining elements of array # if sum > 0, right -= 1
#while left < right: # if sum < 0, left += 1
#find current sum # if it's 0, then add the values to the answer array, and set the left pointer to the next valid value ..
#if sum > 0, right -= 1 # .. (left += 1 while nums[left] == nums[left - 1] && left < right)
#if sum < 0, left += 1 # Return ans[]
#if it's 0, then add the values to the answer array, and set the left pointer to the next valid value ..
#.. (left += 1 while nums[left] == nums[left - 1] && left < right)
#Return ans[]
# @param {Integer[]} nums # @param {Integer[]} nums
# @return {Integer[][]} # @return {Integer[][]}
def three_sum(nums) def three_sum(nums)
#return if length too short # return if length too short
return [] if nums.length < 3 return [] if nums.length < 3
#sort nums, init ans array # sort nums, init ans array
nums, ans = nums.sort, [] nums = nums.sort
ans = []
#loop through nums # loop through nums
nums.each_with_index do |val, ind| nums.each_with_index do |val, ind|
#if the previous value is the same as current, then skip this iteration as it would create duplicates # if the previous value is the same as current, then skip this iteration as it would create duplicates
next if ind > 0 && nums[ind] == nums[ind - 1] next if ind > 0 && nums[ind] == nums[ind - 1]
#init & run two pointer search # init & run two pointer search
left, right = ind + 1, nums.length - 1 left = ind + 1
right = nums.length - 1
while left < right while left < right
#find current sum # find current sum
sum = val + nums[left] + nums[right] sum = val + nums[left] + nums[right]
#decrease sum if it's too great, increase sum if it's too low # decrease sum if it's too great, increase sum if it's too low
if sum > 0 if sum > 0
right -= 1 right -= 1
elsif sum < 0 elsif sum < 0
left += 1 left += 1
#if it's zero, then add the answer to array and set left pointer to next valid value # if it's zero, then add the answer to array and set left pointer to next valid value
else else
ans << [val, nums[left], nums[right]] ans << [val, nums[left], nums[right]]
left += 1 left += 1
while nums[left] == nums[left - 1] && left < right left += 1 while nums[left] == nums[left - 1] && left < right
left += 1
end
end
end end
end end
end
#return answer # return answer
ans ans
end end
nums = [-1, 0, 1, 2, -1, -4]
print three_sum(nums)
# Output: [[-1,-1,2],[-1,0,1]]
nums = []
print three_sum(nums)
# Output: []
nums = [0]
print three_sum(nums)
# Output: []