Sharing an answer code of mine about PassingCars problem of Codility lesson 5.

## Lesson 5: 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, K)
```

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).

Elements of input arrays can be modified.

## Example answer code in Python 2.7

```
def solution(A):
# write your code in Python 2.7
zero_count = 0 # the number of zero occurs
passing_count = 0 # the number of car passing
for i in range(len(A)):
if A[i] == 0:
zero_count = zero_count + 1
else: # if A[i] == 1
passing_count = passing_count + zero_count
if passing_count > 1000000000: # exception point
return -1
return passing_count
```

- Detected time complexity: O(N)