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A One-Pass Hash Table Solution to twoSum


Python Hash Table ImplementationHackerRank Ransom Note, solved using Data.Map as a hash tableHash table using linear probingHackerrank “Hash Tables: Ransom Note” Javascript SolutionDirect address hash tableFour sum algorithm mock interview practiceCounting out game with std::listPython 3 two-sum performanceHash table in PythonHash table solution to twoSum













0












$begingroup$


I read the guiding solution to twoSum in leetcodes ;




Approach 3: One-pass Hash Table

It turns out we can do it in one-pass. While we iterate and inserting elements into the table, we also look back to check if current element's complement already exists in the table. If it exists, we have found a solution and return immediately.




and mimic a python solution



class Solution:
def twoSum(self, nums, target) -> List[int]:
"""
:type nums: List[int]
:type target: int
"""
nums_d =

for i in range(len(nums)):
complement = target - nums[i]

if nums_d.get(complement) != None: #Check None not True Value
return [i, nums_d.get(complement)]
nums_d[nums[i]] = i #produce a map
return []


Unfortunately, my solution is only faster than 81%, which I thought was a best solution.




Runtime: 40 ms, faster than 81.00% of Python3 online submissions for Two Sum.
Memory Usage: 14.3 MB, less than 5.08% of Python3 online submissions for Two Sum.
Next challenges:




How could continue to improve the code, and I am curious about the approaches of the top 20%.










share|improve this question







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.







$endgroup$
















    0












    $begingroup$


    I read the guiding solution to twoSum in leetcodes ;




    Approach 3: One-pass Hash Table

    It turns out we can do it in one-pass. While we iterate and inserting elements into the table, we also look back to check if current element's complement already exists in the table. If it exists, we have found a solution and return immediately.




    and mimic a python solution



    class Solution:
    def twoSum(self, nums, target) -> List[int]:
    """
    :type nums: List[int]
    :type target: int
    """
    nums_d =

    for i in range(len(nums)):
    complement = target - nums[i]

    if nums_d.get(complement) != None: #Check None not True Value
    return [i, nums_d.get(complement)]
    nums_d[nums[i]] = i #produce a map
    return []


    Unfortunately, my solution is only faster than 81%, which I thought was a best solution.




    Runtime: 40 ms, faster than 81.00% of Python3 online submissions for Two Sum.
    Memory Usage: 14.3 MB, less than 5.08% of Python3 online submissions for Two Sum.
    Next challenges:




    How could continue to improve the code, and I am curious about the approaches of the top 20%.










    share|improve this question







    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.







    $endgroup$














      0












      0








      0





      $begingroup$


      I read the guiding solution to twoSum in leetcodes ;




      Approach 3: One-pass Hash Table

      It turns out we can do it in one-pass. While we iterate and inserting elements into the table, we also look back to check if current element's complement already exists in the table. If it exists, we have found a solution and return immediately.




      and mimic a python solution



      class Solution:
      def twoSum(self, nums, target) -> List[int]:
      """
      :type nums: List[int]
      :type target: int
      """
      nums_d =

      for i in range(len(nums)):
      complement = target - nums[i]

      if nums_d.get(complement) != None: #Check None not True Value
      return [i, nums_d.get(complement)]
      nums_d[nums[i]] = i #produce a map
      return []


      Unfortunately, my solution is only faster than 81%, which I thought was a best solution.




      Runtime: 40 ms, faster than 81.00% of Python3 online submissions for Two Sum.
      Memory Usage: 14.3 MB, less than 5.08% of Python3 online submissions for Two Sum.
      Next challenges:




      How could continue to improve the code, and I am curious about the approaches of the top 20%.










      share|improve this question







      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 read the guiding solution to twoSum in leetcodes ;




      Approach 3: One-pass Hash Table

      It turns out we can do it in one-pass. While we iterate and inserting elements into the table, we also look back to check if current element's complement already exists in the table. If it exists, we have found a solution and return immediately.




      and mimic a python solution



      class Solution:
      def twoSum(self, nums, target) -> List[int]:
      """
      :type nums: List[int]
      :type target: int
      """
      nums_d =

      for i in range(len(nums)):
      complement = target - nums[i]

      if nums_d.get(complement) != None: #Check None not True Value
      return [i, nums_d.get(complement)]
      nums_d[nums[i]] = i #produce a map
      return []


      Unfortunately, my solution is only faster than 81%, which I thought was a best solution.




      Runtime: 40 ms, faster than 81.00% of Python3 online submissions for Two Sum.
      Memory Usage: 14.3 MB, less than 5.08% of Python3 online submissions for Two Sum.
      Next challenges:




      How could continue to improve the code, and I am curious about the approaches of the top 20%.







      algorithm python-3.x programming-challenge






      share|improve this question







      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|improve this question







      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|improve this question




      share|improve this question






      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 13 mins ago









      AliceAlice

      1603




      1603




      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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