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A One-Pass Hash Table Solution to twoSum

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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%.

algorithm python-3.x programming-challenge

share

asked 8 mins ago

AliceAlice

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.

$endgroup$

add a comment | 

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%.

algorithm python-3.x programming-challenge

share

asked 8 mins ago

AliceAlice

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.

$endgroup$

add a comment | 

$begingroup$

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%.

algorithm python-3.x programming-challenge

share

asked 8 mins ago

AliceAlice

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.

$endgroup$

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 [] 

share

asked 8 mins ago

AliceAlice

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.

share

asked 8 mins ago

AliceAlice

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.

asked 8 mins ago

AliceAlice

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.

asked 8 mins ago

AliceAlice

1603