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Passing Cars

Task description
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] = 1We 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] = 1the 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.

O(N*2) Solution
------------------------------------------------

class Solution { public int solution(int[] A) { // write your code in Java SE 8 int len=A.length; long count=0; for(int i=0;i<len;i++){ if(A[i]==0){ for(int j=i;j<len;j++){ if(A[j]!=0){ count++; } } } } if(count>1000000000){ count=-1;} return (int) count; } }


Improve the performance wherein O(N) solution is possible.

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