```yes = "y" or "Y"
no = "n" or "N"
quest1 = input("solved problem 1: " + yes + "/" + no + " ")
quest2 = input("solved problem 2: " + yes + "/" + no+ " ")

if quest1 or quest2 == no:
elif quest1 or quest2 == yes:
print("You should not come to class.")
exit()

quest3 = input("solved part 1: " + yes + "/" + no + " ")
quest4 = input("solved part 2: " + yes + "/" + no+ " ")
quest5 = input("solved part 3: " + yes + "/" + no + " ")
quest6 = input("solved part 4: " + yes + "/" + no+ " ")

```

#how can I count these y/n into the number of times of occurrences as output? so I can use if statements to print as per the number of times yes and no is answered.

Oct 13, 2020 in Python
edited Oct 13, 2020 814 views

## 1 answer to this question.

Hi, @Reshma,

You can go through this as an example:

```def count(a, b, m, n):

# If both first and second string

# is empty, or if second string

# is empty, return 1

if ((m == 0 and n == 0) or n == 0):

return 1

# If only first string is empty

# and second string is not empty,

# return 0

if (m == 0):

return 0

# If last characters are same

# Recur for remaining strings by

# 1. considering last characters

#    of both strings

# 2. ignoring last character

#    of first string

if (a[m - 1] == b[n - 1]):

return (count(a, b, m - 1, n - 1) +

count(a, b, m - 1, n))

else:

return count(a, b, m - 1, n)

print(count(a, b, len(a),len(b)))```

Output:

`4`
answered Oct 13, 2020 by Lisa sanyal

+1 vote

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