A set solution to 3 sum which slower than 80%

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A set solution to 3 sum which slower than 80%














1












$begingroup$


I use a dummy solution to solve 3Sum problem.



that is employ set date type to handle duplicates then transform back list.




Given an array nums of n integers, are there elements a, b in nums such that a + b = 9? Find all unique couples in the array which gives the sum of zero.



Note:



The solution set must not contain duplicate triplets.



Example:



Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]



My codes:



from typing import List
import logging
import unittest
import random
from collections import defaultdict,Counter

##logging.disable(level=#logging.CRITICAL)
##logging.basicConfig(level=#logging.DEBUG, format="%(levelname)s %(message)s")

class Solution:
def threeSum(self, nums, target: int=0) -> List[List[int]]:
"""
:type nums: List[int]
:type target: int
"""
if len(nums) < 3: return []

triplets = []
if target == [0, 0, 0]:
triplets.append([0, 0, 0])
return triplets # finish fast

lookup = nums[i]:i for i in range(len(nums)) #overwrite from the high
if len(lookup) == 1:#assert one identical element
keys = [k for k in lookup.keys()]
if keys[0] != 0:
return []
else:
triplets.append([0,0,0])
return triplets



triplets_set = set()

for i in range(len(nums)):
num_1 = nums[i]
sub_target = target - num_1
# #logging.debug(f"level_1_lookup: lookup")

for j in range(i+1, len(nums)):
num_2 = nums[j]
num_3 = sub_target - num_2
k = lookup.get(num_3) #
if k not in None, i, j: #don't reproduce itself
result = [num_1, num_2, num_3]
result.sort()
result = tuple(result)
triplets_set.add(result)

triplets = [list(t) for t in triplets_set]
return triplets


Run and get report




Your runtime beats 28.86 % of python3 submissions.




Could please give hints to improve?









share







New contributor




Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$
















    1












    $begingroup$


    I use a dummy solution to solve 3Sum problem.



    that is employ set date type to handle duplicates then transform back list.




    Given an array nums of n integers, are there elements a, b in nums such that a + b = 9? Find all unique couples in the array which gives the sum of zero.



    Note:



    The solution set must not contain duplicate triplets.



    Example:



    Given array nums = [-1, 0, 1, 2, -1, -4],

    A solution set is:
    [
    [-1, 0, 1],
    [-1, -1, 2]
    ]



    My codes:



    from typing import List
    import logging
    import unittest
    import random
    from collections import defaultdict,Counter

    ##logging.disable(level=#logging.CRITICAL)
    ##logging.basicConfig(level=#logging.DEBUG, format="%(levelname)s %(message)s")

    class Solution:
    def threeSum(self, nums, target: int=0) -> List[List[int]]:
    """
    :type nums: List[int]
    :type target: int
    """
    if len(nums) < 3: return []

    triplets = []
    if target == [0, 0, 0]:
    triplets.append([0, 0, 0])
    return triplets # finish fast

    lookup = nums[i]:i for i in range(len(nums)) #overwrite from the high
    if len(lookup) == 1:#assert one identical element
    keys = [k for k in lookup.keys()]
    if keys[0] != 0:
    return []
    else:
    triplets.append([0,0,0])
    return triplets



    triplets_set = set()

    for i in range(len(nums)):
    num_1 = nums[i]
    sub_target = target - num_1
    # #logging.debug(f"level_1_lookup: lookup")

    for j in range(i+1, len(nums)):
    num_2 = nums[j]
    num_3 = sub_target - num_2
    k = lookup.get(num_3) #
    if k not in None, i, j: #don't reproduce itself
    result = [num_1, num_2, num_3]
    result.sort()
    result = tuple(result)
    triplets_set.add(result)

    triplets = [list(t) for t in triplets_set]
    return triplets


    Run and get report




    Your runtime beats 28.86 % of python3 submissions.




    Could please give hints to improve?









    share







    New contributor




    Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      1












      1








      1





      $begingroup$


      I use a dummy solution to solve 3Sum problem.



      that is employ set date type to handle duplicates then transform back list.




      Given an array nums of n integers, are there elements a, b in nums such that a + b = 9? Find all unique couples in the array which gives the sum of zero.



      Note:



      The solution set must not contain duplicate triplets.



      Example:



      Given array nums = [-1, 0, 1, 2, -1, -4],

      A solution set is:
      [
      [-1, 0, 1],
      [-1, -1, 2]
      ]



      My codes:



      from typing import List
      import logging
      import unittest
      import random
      from collections import defaultdict,Counter

      ##logging.disable(level=#logging.CRITICAL)
      ##logging.basicConfig(level=#logging.DEBUG, format="%(levelname)s %(message)s")

      class Solution:
      def threeSum(self, nums, target: int=0) -> List[List[int]]:
      """
      :type nums: List[int]
      :type target: int
      """
      if len(nums) < 3: return []

      triplets = []
      if target == [0, 0, 0]:
      triplets.append([0, 0, 0])
      return triplets # finish fast

      lookup = nums[i]:i for i in range(len(nums)) #overwrite from the high
      if len(lookup) == 1:#assert one identical element
      keys = [k for k in lookup.keys()]
      if keys[0] != 0:
      return []
      else:
      triplets.append([0,0,0])
      return triplets



      triplets_set = set()

      for i in range(len(nums)):
      num_1 = nums[i]
      sub_target = target - num_1
      # #logging.debug(f"level_1_lookup: lookup")

      for j in range(i+1, len(nums)):
      num_2 = nums[j]
      num_3 = sub_target - num_2
      k = lookup.get(num_3) #
      if k not in None, i, j: #don't reproduce itself
      result = [num_1, num_2, num_3]
      result.sort()
      result = tuple(result)
      triplets_set.add(result)

      triplets = [list(t) for t in triplets_set]
      return triplets


      Run and get report




      Your runtime beats 28.86 % of python3 submissions.




      Could please give hints to improve?









      share







      New contributor




      Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I use a dummy solution to solve 3Sum problem.



      that is employ set date type to handle duplicates then transform back list.




      Given an array nums of n integers, are there elements a, b in nums such that a + b = 9? Find all unique couples in the array which gives the sum of zero.



      Note:



      The solution set must not contain duplicate triplets.



      Example:



      Given array nums = [-1, 0, 1, 2, -1, -4],

      A solution set is:
      [
      [-1, 0, 1],
      [-1, -1, 2]
      ]



      My codes:



      from typing import List
      import logging
      import unittest
      import random
      from collections import defaultdict,Counter

      ##logging.disable(level=#logging.CRITICAL)
      ##logging.basicConfig(level=#logging.DEBUG, format="%(levelname)s %(message)s")

      class Solution:
      def threeSum(self, nums, target: int=0) -> List[List[int]]:
      """
      :type nums: List[int]
      :type target: int
      """
      if len(nums) < 3: return []

      triplets = []
      if target == [0, 0, 0]:
      triplets.append([0, 0, 0])
      return triplets # finish fast

      lookup = nums[i]:i for i in range(len(nums)) #overwrite from the high
      if len(lookup) == 1:#assert one identical element
      keys = [k for k in lookup.keys()]
      if keys[0] != 0:
      return []
      else:
      triplets.append([0,0,0])
      return triplets



      triplets_set = set()

      for i in range(len(nums)):
      num_1 = nums[i]
      sub_target = target - num_1
      # #logging.debug(f"level_1_lookup: lookup")

      for j in range(i+1, len(nums)):
      num_2 = nums[j]
      num_3 = sub_target - num_2
      k = lookup.get(num_3) #
      if k not in None, i, j: #don't reproduce itself
      result = [num_1, num_2, num_3]
      result.sort()
      result = tuple(result)
      triplets_set.add(result)

      triplets = [list(t) for t in triplets_set]
      return triplets


      Run and get report




      Your runtime beats 28.86 % of python3 submissions.




      Could please give hints to improve?







      python algorithm





      share







      New contributor




      Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.










      share







      New contributor




      Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.








      share



      share






      New contributor




      Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 7 mins ago









      AliceAlice

      1764




      1764




      New contributor




      Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.




















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