def solution(arr):
  total = sum(arr)
  mem = [0] * len(arr)
  mem[0] = total
  for i, j in enumerate(arr[1:], start=1):
    mem[i] = mem[i-1] - j

  res = 0
  for i, j in enumerate(arr):
    if j != 0:
      continue
    res += mem[i]
  return res

#Challenge

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 zeroes and/or ones:

$0$ 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 $0\le P<Q<N$, is passing when $P$ is traveling to the east and $Q$ is traveling to the west.

For example with an array A:

A[0] = 0
A[1] = 1
A[2] = 0
A[3] = 1
A[4] = 1

We have five pairs of passing cars: $(0,1),(0,3),(0,4),(2,3),(2,4)$, so your solution should return 5.

#Solution

The thing to notice about this problem is that for each zero we encounter, we add all the ones we encounter in the array after that zero.

In the previous example, there's a zero at index $0$, there's $3$ passing cars ahead of it which we encounter at index $1,3,4$. Similairly we encounter another zero at index $2$ and encounter $2$ passing cars at index $3,4$.

The best way to approach this problem is to keep track of how many cars are ahead of each car at each index. We could do this using a nested loop, running in $\mathcal{O}(n^2)$, but a more efficient approach would be to maintain a decreasing sum for each car index. The first half of my solution above assigns such a sum to the array mem. Beginning with the sum of all values in the array, each index subtracts the corresponding value in A from the index before it.

For the example above this leaves us with:

mem[0] = 3
mem[1] = 2
mem[2] = 2
mem[3] = 1
mem[4] = 0

Now we can just iterate from the start of the array to the end, incrementing a counter by the value of mem[i] for each $i$ where $A[i]==0$. This gives us $3+2=5$ which is the correct answer for our example case.