# codility 5-2. PassingCars

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문제출처

## 문제

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
A = 1
A = 0
A = 1
A = 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
A = 1
A = 0
A = 1
A = 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).


## 풀이코드 - O(n^2)

• Detected time complexity: O(N)
• 조합 가능한 (0,1) pair의 갯수를 찾는다
• 조건
• 0의 index < 1의 index
• pair의 갯수가 100만이 넘으면 -1 리턴
• 시간 복잡도 O(N)
def solution(A):
index_zero = [i for i, x in enumerate(A) if x == 0]
result = 0
for i in index_zero:
result += A[i+1:].count(1)

if result > 1000000000:
return -1
return result



## 다른사람 풀이

• 리스트를 순회하면서, 0의 갯수를 누적해서 더하고, 1을 만나면 누적 값을 result에 합산한다
def solution(A):
result = 0
count_zero = 0
for i in A:
if i == 1 and count_zero == 0:
continue
elif i == 0:
count_zero += 1
elif i == 1:
result += count_zero

if result > 1000000000:
return -1
return result