## Codility : PassingCars

A non-empty zero-indexed array A consisting of N integers is given. The consecutive elements of array A represent consecutive cars on a road.

Array A contains only 0s and/or 1s:

0 represents a car traveling east,

1 represents a car traveling west.

The goal is to count passing cars. We say that a pair of cars (P, Q), where 0 ≤ P < Q < N, is passing when P is traveling to the east and Q is traveling to the west.

For example, consider array A such that:

A[0] = 0

A[1] = 1

A[2] = 0

A[3] = 1

A[4] = 1

We have five pairs of passing cars: (0, 1), (0, 3), (0, 4), (2, 3), (2, 4).

Write a function:

def solution(A)

that, given a non-empty zero-indexed array A of N integers, returns the number of pairs of passing cars.

The function should return −1 if the number of pairs of passing cars exceeds 1,000,000,000.

For example, given:

A[0] = 0

A[1] = 1

A[2] = 0

A[3] = 1

A[4] = 1

the function should return 5, as explained above.

Assume that:

N is an integer within the range [1..100,000];

each element of array A is an integer that can have one of the following values: 0, 1.

Complexity:

expected worst-case time complexity is O(N);

expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

**Python solution : **

def solution(A):

# write your code in Python 2.7

if not A:

return -1

n=len(A)

passed=0

flag1=0

flag2=0

for i in range(0,n) :

flag2=0

if A[i]==0 :

flag1+=1

if A[i]==1 :

flag2=1

if flag2==1 and flag1>0:

passed+=flag1

if passed > 1000000000:

return -1

return passed