تحديات برمجيةالتحدي الثاني - حل التمرين الخامس بلغة بايثون

array = [1, 2, 5, 3, 2, 4, 7, 2]
found = 0

x = int(input('Enter a number: '))

for i in range(0, len(array)):
    if array[i] == x:
        found += 1

if found == 1:
    print("'" + str(x) + "' is found 1 time")
elif found > 1:
    print("'" + str(x) + "' is found " + str(found) + " times")
else:
    print("'" + str(x) + "' is not found ")