Define a function overlapping() that takes two lists and returns True if they have at least one member in common, False otherwise.
- You may use your is_member() function, or the in operator.
- But for the sake of the exercise, you should (also) write it using two nested for-loops.
I mainly compare Python's in-built set functions against 'standard' looping mechanisms.
The results of all 3 solutions are the same. Only the implementation differs.
Solution #1 uses the Set's .intersection()
method, to find common elements between two lists.
Solution #2 loops against two lists, and maintains another list of common elements to check against.
Solution #3 loops against one list only. It breaks immediately when a common member is found. (Reason: the condition is to "have at least one member in common", and not anything more complex. ) This is possibly a performant solution, but also one likely to introduce bugs.
I'd go with Solution #1 - it appears cleanest. Also, due to the use of sets, is perhaps more performant than Solution #3.
def overlapping_1(l1, l2) -> bool:
if set(l1).intersection(l2):
return True
else:
return False
def overlapping_2(l1, l2) -> bool:
common = []
for i in l1:
for j in l2:
if i == j:
common.append(i)
if common != []:
return True
else:
return False
def overlapping_3(l1, l2) -> bool:
for i in l1:
if i in l2:
return True # Immediately exit
else:
return False
if __name__ == "__main__":
test_l1 = ["a", "b", "c"]
test_l2 = ["c", "d", "e"]
test_l3 = [1, 2, 3]
funcs = overlapping_1, overlapping_2, overlapping_3
for f in funcs:
print(f(test_l1, test_l2), test_l1, test_l2)
print(f(test_l1, test_l3), test_l1, test_l3)