Big O notation compare with itertools.product and for loop
I have two array of integers and I want to the minimum absolute difference between any elements from two array. I have two different way to do it. Both seems pretty much the same, but I do not understand what is Big O notitaion for two methods.
are they both are O(mk)?
list1 = [1,3,7,12]
list2 = [11,39,9,-1]
from itertools import product
def method1(l1,l2):
return min([abs(x[0]-x[1]) for x in product(l1, l2)])
def method2(l1,l2):
num =
for i in l1:
for j in l2:
num.append(abs(i-j))
return min(num)
print(method1(list1,list2))
print(method2(list1,list2))
python python-3.x
asked Nov 22 '18 at 2:13
jacobcan118jacobcan118
640423
add a comment |
I have two array of integers and I want to the minimum absolute difference between any elements from two array. I have two different way to do it. Both seems pretty much the same, but I do not understand what is Big O notitaion for two methods.
are they both are O(mk)?
list1 = [1,3,7,12]
list2 = [11,39,9,-1]
from itertools import product
def method1(l1,l2):
return min([abs(x[0]-x[1]) for x in product(l1, l2)])
def method2(l1,l2):
num =
for i in l1:
for j in l2:
num.append(abs(i-j))
return min(num)
print(method1(list1,list2))
print(method2(list1,list2))
python python-3.x
asked Nov 22 '18 at 2:13
jacobcan118jacobcan118
640423
its the same time complexity
– juanpa.arrivillaga
Nov 22 '18 at 2:29
add a comment |
I have two array of integers and I want to the minimum absolute difference between any elements from two array. I have two different way to do it. Both seems pretty much the same, but I do not understand what is Big O notitaion for two methods.
are they both are O(mk)?
list1 = [1,3,7,12]
list2 = [11,39,9,-1]
from itertools import product
def method1(l1,l2):
return min([abs(x[0]-x[1]) for x in product(l1, l2)])
def method2(l1,l2):
num =
for i in l1:
for j in l2:
num.append(abs(i-j))
return min(num)
print(method1(list1,list2))
print(method2(list1,list2))
python python-3.x
asked Nov 22 '18 at 2:13
jacobcan118jacobcan118
640423
I have two array of integers and I want to the minimum absolute difference between any elements from two array. I have two different way to do it. Both seems pretty much the same, but I do not understand what is Big O notitaion for two methods.
are they both are O(mk)?
list1 = [1,3,7,12]
list2 = [11,39,9,-1]
from itertools import product
def method1(l1,l2):
return min([abs(x[0]-x[1]) for x in product(l1, l2)])
def method2(l1,l2):
num =
for i in l1:
for j in l2:
num.append(abs(i-j))
return min(num)
print(method1(list1,list2))
print(method2(list1,list2))
python python-3.x
python python-3.x
asked Nov 22 '18 at 2:13
jacobcan118jacobcan118
640423
asked Nov 22 '18 at 2:13
jacobcan118jacobcan118
640423
asked Nov 22 '18 at 2:13
jacobcan118jacobcan118
640423
asked Nov 22 '18 at 2:13
jacobcan118jacobcan118
640423
asked Nov 22 '18 at 2:13
jacobcan118jacobcan118
640423
640423
its the same time complexity
– juanpa.arrivillaga
Nov 22 '18 at 2:29
add a comment |
its the same time complexity
– juanpa.arrivillaga
Nov 22 '18 at 2:29
its the same time complexity
– juanpa.arrivillaga
Nov 22 '18 at 2:29
its the same time complexity
– juanpa.arrivillaga
Nov 22 '18 at 2:29
add a comment |
1 Answer
1
active
oldest
votes
Yes. Both methods are exaclty the same
Suggestion for method1:
def method1(l1,l2):
return min([abs(a - b) for a, b in product(l1, l2)])
answered Nov 22 '18 at 2:49
AResemAResem
1114
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes. Both methods are exaclty the same
Suggestion for method1:
def method1(l1,l2):
return min([abs(a - b) for a, b in product(l1, l2)])
answered Nov 22 '18 at 2:49
AResemAResem
1114
add a comment |
Yes. Both methods are exaclty the same
Suggestion for method1:
def method1(l1,l2):
return min([abs(a - b) for a, b in product(l1, l2)])
answered Nov 22 '18 at 2:49
AResemAResem
1114
add a comment |
Yes. Both methods are exaclty the same
Suggestion for method1:
def method1(l1,l2):
return min([abs(a - b) for a, b in product(l1, l2)])
answered Nov 22 '18 at 2:49
AResemAResem
1114
Yes. Both methods are exaclty the same
Suggestion for method1:
def method1(l1,l2):
return min([abs(a - b) for a, b in product(l1, l2)])
answered Nov 22 '18 at 2:49
AResemAResem
1114
answered Nov 22 '18 at 2:49
AResemAResem
1114
answered Nov 22 '18 at 2:49
AResemAResem
1114
answered Nov 22 '18 at 2:49
AResemAResem
1114
1114
add a comment |
add a comment |
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its the same time complexity
– juanpa.arrivillaga
Nov 22 '18 at 2:29