A One-Pass Hash 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
asked 17 mins ago
AliceAlice
1603
New contributor
Alice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
asked 17 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
asked 17 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.
$endgroup$
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
asked 17 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 17 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 17 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 17 mins ago
AliceAlice
1603
asked 17 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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