CODILITY - LESSON 3 - passingCars

PROBLEM:

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).

MY SOLUTION (100% score, time complexity O(N))
class Solution { public int solution(int[] A) { int l = A.Length; int countZero =0; int countPairs=0; int exceed = 1000000000; for (int i=0; i<l; i++) { if (A[i]==1) countPairs += countZero; else countZero++; } if (countPairs > exceed || countPairs <0) return -1; else return countPairs; } }


The key is to remember that the pairs needs to be all (0,1) as 0 passes before 1 in time. So pairs can't combine (1,0).

Reading the array from position A[0] to A[N] each zero will combine with subsequent 1s only.

PR