[Codility] Demo equilibrium index

A zero-indexed array A consisting of N integers is given. An equilibrium index of this array is any integer P such that 0 ≤ P < N and the sum of elements of lower indices is equal to the sum of elements of higher indices, i.e.
A[0] + A[1] + ... + A[P−1] = A[P+1] + ... + A[N−2] + A[N−1].

Sum of zero elements is assumed to be equal to 0. This can happen if P = 0 or if P = N−1.
For example, consider the following array A consisting of N = 8 elements:
A[0] = -1 A[1] = 3 A[2] = -4 A[3] = 5 A[4] = 1 A[5] = -6 A[6] = 2 A[7] = 1
P = 1 is an equilibrium index of this array, because:
A[0] = −1 = A[2] + A[3] + A[4] + A[5] + A[6] + A[7]

P = 3 is an equilibrium index of this array, because:
A[0] + A[1] + A[2] = −2 = A[4] + A[5] + A[6] + A[7]

P = 7 is also an equilibrium index, because:
A[0] + A[1] + A[2] + A[3] + A[4] + A[5] + A[6] = 0

and there are no elements with indices greater than 7.
P = 8 is not an equilibrium index, because it does not fulfill the condition 0 ≤ P < N.
Write a function:
int solution(int A[], int N);

that, given a zero-indexed array A consisting of N integers, returns any of its equilibrium indices. The function should return −1 if no equilibrium index exists.
For example, given array A shown above, the function may return 1, 3 or 7, as explained above.
Assume that:
N is an integer within the range [0..100,000];
each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

Complexity:
expected worst-case time complexity is O(N);
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

---

public func solution(arr:[Int])->Int{

    guard arr.count > 0 else { return -1 }
    
    let count = arr.count
    
    var totalSum = 0
    var leftSum = 0
    var rightSum = 0
    
    for i in 0 ..< count { totalSum += arr[i]}
    
    for j in 0 ..< count {
        rightSum = totalSum - leftSum - arr[j]
        
        if leftSum == rightSum { return j }
        else { leftSum += arr[j] }
    }
    return -1
}

var arr = [-7,1,5,2,-4,3,0]

solution(arr)