A set solution to 3 sum which slower than 80%
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A set solution to 3 sum which slower than 80%
$begingroup$
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 innums
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 algorithm
New contributor
$endgroup$
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 innums
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 algorithm
New contributor
$endgroup$
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 innums
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 algorithm
New contributor
$endgroup$
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 innums
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 algorithm
python algorithm
New contributor
New contributor
New contributor
asked 7 mins ago
AliceAlice
1764
1764
New contributor
New contributor
add a comment |
add a comment |
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