Refactor a hash table solution to twoSum
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Refactor a hash table solution to twoSum
$begingroup$
I try the most to solve a twoSum problem in leetcode
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
The plan:
1) brute force to iterate len(nums) O(n)
2) search for target - num[i] with a hash table O(1)
Implement
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
nums_d =
for i in range(len(nums)):
nums_d.setdefault(nums[i], []).append(i)
for i in range(len(nums)):
sub_target = target - nums[i]
nums_d[nums[i]].pop(0) #remove the fixer
result = nums_d.get(sub_target)#hash table to search
if result:
return [i, result[0]]
return []
I strives hours for this solution but found that answer accepted but not passed Score 60.
Runtime: 60 ms, faster than 46.66% of Python3 online submissions for Two Sum.
Memory Usage: 16.1 MB, less than 5.08% of Python3 online submissions for Two Sum.
I want to refactor the codes so that to achieve at least faster than 60%.
Could you please provide hints?
algorithm
asked 3 mins ago
AliceAlice
101
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 try the most to solve a twoSum problem in leetcode
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
The plan:
1) brute force to iterate len(nums) O(n)
2) search for target - num[i] with a hash table O(1)
Implement
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
nums_d =
for i in range(len(nums)):
nums_d.setdefault(nums[i], []).append(i)
for i in range(len(nums)):
sub_target = target - nums[i]
nums_d[nums[i]].pop(0) #remove the fixer
result = nums_d.get(sub_target)#hash table to search
if result:
return [i, result[0]]
return []
I strives hours for this solution but found that answer accepted but not passed Score 60.
Runtime: 60 ms, faster than 46.66% of Python3 online submissions for Two Sum.
Memory Usage: 16.1 MB, less than 5.08% of Python3 online submissions for Two Sum.
I want to refactor the codes so that to achieve at least faster than 60%.
Could you please provide hints?
algorithm
asked 3 mins ago
AliceAlice
101
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 try the most to solve a twoSum problem in leetcode
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
The plan:
1) brute force to iterate len(nums) O(n)
2) search for target - num[i] with a hash table O(1)
Implement
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
nums_d =
for i in range(len(nums)):
nums_d.setdefault(nums[i], []).append(i)
for i in range(len(nums)):
sub_target = target - nums[i]
nums_d[nums[i]].pop(0) #remove the fixer
result = nums_d.get(sub_target)#hash table to search
if result:
return [i, result[0]]
return []
I strives hours for this solution but found that answer accepted but not passed Score 60.
Runtime: 60 ms, faster than 46.66% of Python3 online submissions for Two Sum.
Memory Usage: 16.1 MB, less than 5.08% of Python3 online submissions for Two Sum.
I want to refactor the codes so that to achieve at least faster than 60%.
Could you please provide hints?
algorithm
asked 3 mins ago
AliceAlice
101
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 try the most to solve a twoSum problem in leetcode
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
The plan:
1) brute force to iterate len(nums) O(n)
2) search for target - num[i] with a hash table O(1)
Implement
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
nums_d =
for i in range(len(nums)):
nums_d.setdefault(nums[i], []).append(i)
for i in range(len(nums)):
sub_target = target - nums[i]
nums_d[nums[i]].pop(0) #remove the fixer
result = nums_d.get(sub_target)#hash table to search
if result:
return [i, result[0]]
return []
I strives hours for this solution but found that answer accepted but not passed Score 60.
Runtime: 60 ms, faster than 46.66% of Python3 online submissions for Two Sum.
Memory Usage: 16.1 MB, less than 5.08% of Python3 online submissions for Two Sum.
I want to refactor the codes so that to achieve at least faster than 60%.
Could you please provide hints?
algorithm
algorithm
asked 3 mins ago
AliceAlice
101
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 3 mins ago
AliceAlice
101
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 3 mins ago
AliceAlice
101
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 3 mins ago
AliceAlice
101
asked 3 mins ago
AliceAlice
101
101
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.
add a comment |
add a comment |
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