[Codility] Lesson 5 - PassingCars

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:

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.

Source code

#include <algorithm>

int solution(vector<int> &A) {
    int oneCnt = 0;
    int answer = 0;
    for(int i=0;i<A.size();i++) {
        if(A[i] == 1) oneCnt++;
    }

    for(int i=0;i<A.size();i++) {
        if(A[i] == 0) {
            answer += oneCnt;
            if(answer > 1000000000 ) return -1;
        }
        else {
            oneCnt--;
        }
    }
    return answer;
}