[Lesson 5] PassingCars ( score : 90 )

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:
int solution(vector<int> &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.

 

풀이

첫번째 제출 score : 90

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(vector<int> &A) {
    // Implement your solution here
    int size = A.size();
    int sum[100003];
    int count = 0;

    //prefix sum
    sum[0] = A[0];
    for(int i = 1; i < size; i++)
    {
        if(A[i] == 1)
        {
            sum[i] += (sum[i-1] + 1);
        }
        else
        {
            sum[i] += sum[i-1];
        }
    }

    //counting
    for(int i = 0; i < size; i++)
    {
        if(count > 1000000000) return -1;
        if(A[i] == 0)
        {
            count += (sum[size-1] - sum[i]);
        }
    }

    return count;


}

극단적인 상황에서 -1 을 반환하는데 이유를 모르겠다ㅠㅜ.