Lesson 5 Prefix Sums PassingCars 나만의 풀이

문제

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.

대충 해석하자면 A는 N개의 정수가 주어진다.

0부터 셈을 시작해서 1인 부분을 카운팅 하는것이다.

그리고 그 다음 0부터 1인 부분을 카운팅 .... 반복

그리고 총합을 리턴

만약 10억이 넘었다면 -1을 리턴

 

 

첫 번째 시도

첫번째 시도에 많은 생각을 했다. 어떻게 해도 생각한 대로 안 나오길래 밥 먹고 와서 바로 reverse()를 적용하여 풀었다

역시 밥심이야... 

def solution(A):
    index_list = []
    A.reverse()

    count = 0
    for i in range(0, len(A)):
        if A[i]==1:
            count +=1
        else:
            index_list.append(count)
            
    val = sum(index_list)
    if val>1000000000:
        return -1
    return val

 

성공