Second Solution

[Mandraxed]

The following code works for snaplist.

def snaplist(a):
for i in range(0, len(a)-1):
if a[i] == a[i+1]:
print("Yes")
return
print("No")
a = [3, 4, 4, 8]
snaplist(a)

Your code has:
for i in range(0, len(a)-1-1) and does not work for repeated integers at the end of lists

[Mandraxed]

This one looks tricky. You have to answer detailing the output from the code. Perhaps we can look at it next Wednesday if you have time.

class C1:
def init(self, x):
self._x = x
print("C1")

def decreasing(self, a):
    if a < self._x:
        return str(a) + " < " + str(self._x)
    else:
        raise ValueError("Value %d is not not less than %d" % (a, self._x))

class C2(C1):

def __init__(self, x):
    super().__init__(3 * x)
    print("C2")

def decreasing(self, a):
    return super().decreasing(2*a)

for j in [10,40]:
try:
print(j)
a1 = C1(j)
print(a1.decreasing(20))
a2 = C2(20)
print(a2.decreasing(20))
except:
print("problem")
finally:
print("finally")

[Mandraxed]

Came up with a version of Snaplist (Q2(a)) which works as intended, printing 'YES' once if the list contains consecutive elements and printing "NO" once if the list does not contain consecutive elements.

def snaplist(a):
for i in range(0, len(a)-1):
if ((a)[i] == (a)[i + 1]) == True:
print("Yes")
return
else:
print("No")
return

a = [1, 5, 6, 7, 8, 9, 9]
b = [1, 4, 3, 4]

snaplist(a)

[Mandraxed]

Here is a shorter version of bingolist:

def bingolist(a):
for i in range(0, len(a)-1):
for j in range(i + 1, len(a)):
if ((a)[i] == (a)[j]) == True:
print("Yes")
return
if ((a)[i] == (a)[j]) == False:
print("No")
return