Write a function is_present_in_opposite_halves(elem, l1: list, l2: list) that checks whether the given elem is present in opposite halves of the two lists l1 and l2.

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Write a function is_present_in_opposite_halves(elem, l1: list, l2: list) that checks whether the given elem is present in opposite halves of the two lists l1 and l2.
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Got it ✅ You want a function that checks whether a given element elem is present in opposite halves of two even-length lists l1 and l2.

Here’s the correct Python implementation:

def is_present_in_opposite_halves(elem, l1: list, l2: list) -> bool:
    # Find middle indices
    mid1, mid2 = len(l1) // 2, len(l2) // 2

    # Split the lists into halves
    first_half_l1, second_half_l1 = l1[:mid1], l1[mid1:]
    first_half_l2, second_half_l2 = l2[:mid2], l2[mid2:]

    # Case 1: elem in first half of l1 and second half of l2
    if elem in first_half_l1 and elem in second_half_l2:
        return True
    
    # Case 2: elem in second half of l1 and first half of l2
    if elem in second_half_l1 and elem in first_half_l2:
        return True

    # If neither condition is satisfied
    return False

✅ Explanation:
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We split each list into two halves using slicing.
Check two valid cases:
- elem is in the first half of l1 and in the second half of l2
- elem is in the second half of l1 and in the first half of l2
Return True if either case is satisfied, else False.

🔍 Examples:
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print(is_present_in_opposite_halves(3, [1, 2, 3, 4], [5, 6, 3, 8]))  # False
print(is_present_in_opposite_halves(7, [1, 2, 3, 4], [5, 6, 3, 8]))  # False
print(is_present_in_opposite_halves(6, [5, 6, 7, 8], [1, 2, 6, 4]))  # True
print(is_present_in_opposite_halves(6, [5, 7, 6, 8], [1, 6, 2, 4]))  # True

Would you like me to also handle the case where the same element appears in both halves of the same list (so we only check exact opposite halves once)?

OOPE-2 - This article is part of a series.

Part : Python OPPE-2