Why the Output gives 3?

Question

def fun(a,p=0):
if(p==len(a)):
return 0
z=1 if not a[p] else 0
return z + fun(a,p+1)
a=[1,0,1,1,0,1,0]
print(fun(a))

Answer 1

Mahmuda Akter Moon

We are counting 0 so here total number of 0 is 3 in list

See this line

z = 1 if not a[p] else 0

Here if not 1 menas 0 then count is 1 otherwise count is 0 so there is total number of 0 is 3

Answer 2

The function fun will be recursed until the variable 'p' equals 7.
During this, the value of the variable 'z' will change to 0 or 1 depending on the values ​​of the array 'a' (if a[p]==1: z=0 else z=1)
As a result of execution of 5, the line will look like this: "return 0+1+0+0+1+0+1"

Answer 3

z=1 if not a[p] else 0
flips 0 to 1 and 1 to zero.

p+1 in every recursion p becomes +1 greater until the end of the list a and there becomes 0 again. recursion break.
and if to flip 0 to 1 and 1 to zero here will be 3 ones in a list.
so in the end z = 1+1+1==3 or in short:

z = 0+1+0+0+1+0+1

Answer 4

Mahmuda Akter Moon

In 5th line we are adding count like:

0 + 1 + 0 + 0 + 1 + 0 + 1 = 3

That is recursion. We are calling function and incrementing p and checking next value to count 0 till p == len(a)

Answer 5

What about the 5th line?