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A set solution to 3-sum which is slower than 80% of submissions

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1

$begingroup$

I use a dummy solution to solve 3Sum problem.

that is employ set date type to handle duplicates then transform back list.

Given an array nums of n integers, are there elements a, b in nums such that a + b = 9? Find all unique couples in the array which gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:

[
 [-1, 0, 1],
 [-1, -1, 2]
]

My codes:

from typing import List
import logging
import unittest
import random
from collections import defaultdict,Counter

##logging.disable(level=#logging.CRITICAL)
##logging.basicConfig(level=#logging.DEBUG, format="%(levelname)s %(message)s")

 class Solution:
 def threeSum(self, nums, target: int=0) -> List[List[int]]:
 """
 :type nums: List[int]
 :type target: int 
 """
 if len(nums) < 3: return []

 triplets = []
 if target == [0, 0, 0]: 
 triplets.append([0, 0, 0])
 return triplets # finish fast 

 lookup = nums[i]:i for i in range(len(nums)) #overwrite from the high
 if len(lookup) == 1:#assert one identical element 
 keys = [k for k in lookup.keys()]
 if keys[0] != 0: 
 return []
 else:
 triplets.append([0,0,0])
 return triplets



 triplets_set = set()

 for i in range(len(nums)):
 num_1 = nums[i]
 sub_target = target - num_1
 # #logging.debug(f"level_1_lookup: lookup")

 for j in range(i+1, len(nums)):
 num_2 = nums[j]
 num_3 = sub_target - num_2 
 k = lookup.get(num_3) # 
 if k not in None, i, j: #don't reproduce itself 
 result = [num_1, num_2, num_3]
 result.sort()
 result = tuple(result)
 triplets_set.add(result)

 triplets = [list(t) for t in triplets_set]
 return triplets 

Run and get report

Your runtime beats 28.86 % of python3 submissions.

Could please give hints to improve?

python performance algorithm k-sum

share|improve this question

edited 2 mins ago

200_success

130k17155419

asked 2 hours ago

AliceAlice

1764

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.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you check the problem description? It is about pairs (a, b) which sum to 9, but the examples are about triples which sum to zero. – Btw, if you beat 28.86 of other submissions then you are slower 71.14 percent, not slower than 80 percent :)
  $endgroup$
  – Martin R
  4 mins ago
  
  

  
  
  
  

add a comment | 

1

$begingroup$

I use a dummy solution to solve 3Sum problem.

that is employ set date type to handle duplicates then transform back list.

Given an array nums of n integers, are there elements a, b in nums such that a + b = 9? Find all unique couples in the array which gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:

[
 [-1, 0, 1],
 [-1, -1, 2]
]

My codes:

from typing import List
import logging
import unittest
import random
from collections import defaultdict,Counter

##logging.disable(level=#logging.CRITICAL)
##logging.basicConfig(level=#logging.DEBUG, format="%(levelname)s %(message)s")

 class Solution:
 def threeSum(self, nums, target: int=0) -> List[List[int]]:
 """
 :type nums: List[int]
 :type target: int 
 """
 if len(nums) < 3: return []

 triplets = []
 if target == [0, 0, 0]: 
 triplets.append([0, 0, 0])
 return triplets # finish fast 

 lookup = nums[i]:i for i in range(len(nums)) #overwrite from the high
 if len(lookup) == 1:#assert one identical element 
 keys = [k for k in lookup.keys()]
 if keys[0] != 0: 
 return []
 else:
 triplets.append([0,0,0])
 return triplets



 triplets_set = set()

 for i in range(len(nums)):
 num_1 = nums[i]
 sub_target = target - num_1
 # #logging.debug(f"level_1_lookup: lookup")

 for j in range(i+1, len(nums)):
 num_2 = nums[j]
 num_3 = sub_target - num_2 
 k = lookup.get(num_3) # 
 if k not in None, i, j: #don't reproduce itself 
 result = [num_1, num_2, num_3]
 result.sort()
 result = tuple(result)
 triplets_set.add(result)

 triplets = [list(t) for t in triplets_set]
 return triplets 

Run and get report

Your runtime beats 28.86 % of python3 submissions.

Could please give hints to improve?

python performance algorithm k-sum

share|improve this question

edited 2 mins ago

200_success

130k17155419

asked 2 hours ago

AliceAlice

1764

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.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you check the problem description? It is about pairs (a, b) which sum to 9, but the examples are about triples which sum to zero. – Btw, if you beat 28.86 of other submissions then you are slower 71.14 percent, not slower than 80 percent :)
  $endgroup$
  – Martin R
  4 mins ago
  
  

  
  
  
  

add a comment | 

$begingroup$

I use a dummy solution to solve 3Sum problem.

that is employ set date type to handle duplicates then transform back list.

Given an array nums of n integers, are there elements a, b in nums such that a + b = 9? Find all unique couples in the array which gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:

[
 [-1, 0, 1],
 [-1, -1, 2]
]

My codes:

from typing import List
import logging
import unittest
import random
from collections import defaultdict,Counter

##logging.disable(level=#logging.CRITICAL)
##logging.basicConfig(level=#logging.DEBUG, format="%(levelname)s %(message)s")

 class Solution:
 def threeSum(self, nums, target: int=0) -> List[List[int]]:
 """
 :type nums: List[int]
 :type target: int 
 """
 if len(nums) < 3: return []

 triplets = []
 if target == [0, 0, 0]: 
 triplets.append([0, 0, 0])
 return triplets # finish fast 

 lookup = nums[i]:i for i in range(len(nums)) #overwrite from the high
 if len(lookup) == 1:#assert one identical element 
 keys = [k for k in lookup.keys()]
 if keys[0] != 0: 
 return []
 else:
 triplets.append([0,0,0])
 return triplets



 triplets_set = set()

 for i in range(len(nums)):
 num_1 = nums[i]
 sub_target = target - num_1
 # #logging.debug(f"level_1_lookup: lookup")

 for j in range(i+1, len(nums)):
 num_2 = nums[j]
 num_3 = sub_target - num_2 
 k = lookup.get(num_3) # 
 if k not in None, i, j: #don't reproduce itself 
 result = [num_1, num_2, num_3]
 result.sort()
 result = tuple(result)
 triplets_set.add(result)

 triplets = [list(t) for t in triplets_set]
 return triplets 

Run and get report

Your runtime beats 28.86 % of python3 submissions.

Could please give hints to improve?

python performance algorithm k-sum

share|improve this question

edited 2 mins ago

200_success

130k17155419

asked 2 hours ago

AliceAlice

1764

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.

$endgroup$

from typing import List
import logging
import unittest
import random
from collections import defaultdict,Counter

##logging.disable(level=#logging.CRITICAL)
##logging.basicConfig(level=#logging.DEBUG, format="%(levelname)s %(message)s")

 class Solution:
 def threeSum(self, nums, target: int=0) -> List[List[int]]:
 """
 :type nums: List[int]
 :type target: int 
 """
 if len(nums) < 3: return []

 triplets = []
 if target == [0, 0, 0]: 
 triplets.append([0, 0, 0])
 return triplets # finish fast 

 lookup = nums[i]:i for i in range(len(nums)) #overwrite from the high
 if len(lookup) == 1:#assert one identical element 
 keys = [k for k in lookup.keys()]
 if keys[0] != 0: 
 return []
 else:
 triplets.append([0,0,0])
 return triplets



 triplets_set = set()

 for i in range(len(nums)):
 num_1 = nums[i]
 sub_target = target - num_1
 # #logging.debug(f"level_1_lookup: lookup")

 for j in range(i+1, len(nums)):
 num_2 = nums[j]
 num_3 = sub_target - num_2 
 k = lookup.get(num_3) # 
 if k not in None, i, j: #don't reproduce itself 
 result = [num_1, num_2, num_3]
 result.sort()
 result = tuple(result)
 triplets_set.add(result)

 triplets = [list(t) for t in triplets_set]
 return triplets 

share|improve this question

edited 2 mins ago

200_success

130k17155419

asked 2 hours ago

AliceAlice

1764

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|improve this question

edited 2 mins ago

200_success

130k17155419

asked 2 hours ago

AliceAlice

1764

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.

edited 2 mins ago

200_success

130k17155419

edited 2 mins ago

200_success

130k17155419

asked 2 hours ago

AliceAlice

1764

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 2 hours ago

AliceAlice

1764