A One-Pass Hash Table Solution to twoSumPython 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
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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
$begingroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
algorithm python-3.x programming-challenge
New contributor
New contributor
New contributor
asked 13 mins ago
AliceAlice
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1603
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