# Find the number of consecutive zero at the end after multiplying n numbers in Python

Suppose we have an array with n numbers, we have to return the number of consecutive zero’s at the end after multiplying all the n numbers.

So, if the input is like [200, 20, 5, 30, 40, 14], then the output will be 6 as 200 * 20 * 5 * 30 * 40 * 14 = 336000000, there are six 0s at the end.

To solve this, we will follow these steps −

• Define a function count_fact_two() . This will take n

• count := 0

• while n mod 2 is 0, do

• count := count + 1

• n := n / 2 (only the quotient as integer)

• return count

• Define a function count_fact_five() . This will take n

• count := 0

• while n mod 5 is 0, do

• count := count + 1

• n := n / 5 (only the quotient as integer)

• return count

• From the main method, do the following −

• n := size of A

• twos := 0, fives := 0

• for i in range 0 to n, do

• twos := twos + count_fact_two(A[i])

• fives := fives + count_fact_five(A[i])

• if twos − fives, then

• return twos

• otherwise,

• return fives

## Example

Let us see the following implementation to get better understanding −

Live Demo

def count_fact_two( n ):
count = 0
while n % 2 == 0:
count+=1
n = n // 2
return count
def count_fact_five( n ):
count = 0
while n % 5 == 0:
count += 1
n = n // 5
return count
def get_consecutive_zeros(A):
n = len(A)
twos = 0
fives = 0
for i in range(n):
twos += count_fact_two(A[i])
fives += count_fact_five(A[i])
if twos < fives:
return twos
else:
return fives
A = [200, 20, 5, 30, 40, 14]
print(get_consecutive_zeros(A))

## Input

[200, 20, 5, 30, 40, 14]

## Output

6