Codility 5.1 Passing Cars

Task description
A non-empty 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 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.

Write an efficient algorithm for the following assumptions:

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.

解题思路：每发现一个1，这个1前面的0有几个，pairs就增加几个。

def solution(A):
    # write your code in Python 3.6
    re = 0
    count_zero = 0
    for i in range(0, len(A)):
        if A[i] == 0:
            count_zero += 1
        elif A[i] == 1:
            re += count_zero
            if re > 1000000000:
                return -1
    return re