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Refactor a hash table solution to twoSum

0

$begingroup$

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

share

asked 54 secs 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 | 

0

$begingroup$

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

share

asked 54 secs 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

share

asked 54 secs 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.

$endgroup$

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 []

share

asked 54 secs 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.

share

asked 54 secs 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 54 secs 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 54 secs ago

AliceAlice

101