Yes you should find the equal squared elements one to one as you put it.
For example:
a1=[2,2,3]
a2=[4,9,4]
This would return true, as there are two "2" elements and one "3" in a1 array and when squared will be equal to a2 irrespective of their position.
a1[4,4,5]
a2[16,25,25]
This would return false.
Hopes this clears up your confusion.
got it! thx!
Yes you should find the equal squared elements one to one as you put it.
For example:
a1=[2,2,3]
a2=[4,9,4]
This would return true, as there are two "2" elements and one "3" in a1 array and when squared will be equal to a2 irrespective of their position.
a1[4,4,5]
a2[16,25,25]
This would return false.
Hopes this clears up your confusion.
This comment is hidden because it contains spoiler information about the solution
Step by step guide:
2**2 != 9a1 = [2, 2, 3]
a2 = [4, 9, 9]
test.assert_equals(comp(a1, a2), False)
Why this case's result is not True?