Codility - Lesson 5 Prefix Sums - 1. PassingCars

Source Link:
https://codility.com/programmers/lessons/5-prefix_sums/passing_cars/

Question:
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] = 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:

class Solution { public int solution(int[] 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] = 0 A[1] = 1 A[2] = 0 A[3] = 1 A[4] = 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).

Elements of input arrays can be modified.

My Solution:
Notes:
1. find number of pairs (P, Q), where P < Q (important)
// try to use "one pass" solution (low time complexity)
2. just look back "num_east", that will be the number of cars can be paired (with the current car)
for(int i=0; i<A.length; i++){
if(A[i] ==0){ // to east
num_east++;
}
if(A[i] ==1){ // to west
num_pass = num_pass + num_east;
}
}  
3. "num_pass < 0" is for the "overflow" cases, when "overflow" occurs, the value will "< 0" (important)
// can use "_" for a big value
if(num_pass > 1_000_000_000 || num_pass < 0) 
return -1;