Problem

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.

Discussion

1. I mainly compare Python's in-built set functions against 'standard' looping mechanisms.

2. The results of all 3 solutions are the same. Only the implementation differs.

3. Solution #1 uses the Set's .intersection() method, to find common elements between two lists.

4. Solution #2 loops against two lists, and maintains another list of common elements to check against.

5. 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.

6. I'd go with Solution #1 - it appears cleanest. Also, due to the use of sets, is perhaps more performant than Solution #3.

Solution

    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)