تحديات برمجيةالتحدي الثالث - حل التمرين الثاني بلغة بايثون

# هنا قمنا بتعريف الدالة
def count_occurrence(arr):

    for i in range(0, len(arr)):

        counter = 1

        if arr[i] is not None:

            for j in range(i + 1, len(arr)):

                if arr[i] == arr[j]:
                    counter += 1
                    arr[j] = None

            print("[" + str(arr[i]) + "] is repeated " + str(counter) + " time(s)")


# هنا قمنا بتجهيز القيم التي سنمررها للدالة
array = [0, 4, 2, 3, 2, 4, 3, 5, 2, 0, 1, 4, 2]

# هنا قمنا بتجربة الدالة
count_occurrence(array)

[0] is repeated 2 time(s)
[4] is repeated 3 time(s)
[2] is repeated 4 time(s)
[3] is repeated 2 time(s)
[5] is repeated 1 time(s)
[1] is repeated 1 time(s)