## miércoles, 3 de agosto de 2016

### Equilibrium Index of Array

I was waiting for do a test for a job and they send me a codility test.  They have a demo task about the equilibrium of array.

"

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:
function solution(A);
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.
"

$$\sum_{\substack{ 0\le i\lt N}} A_i=k$$

Suppouse that p is an equilibrium index, then

$$\text{ (1) }\sum_{\substack{0\le i\lt P}}A_i+A_p+\sum_{\substack{P\lt i\lt N}}A_i=k$$
Where
$$\text{ (2) }\sum_{\substack{0\le i\lt P}}A_i=\sum_{\substack{P\lt i\lt N}}A_i$$
From (1) and (2) we can conclude that:
$$\text{ (3) }2* \sum_{\substack{0\le i\lt P}}A_i+A_p=k$$
We can rewrite (3) and then we can conclude that p is an Equilibrium index if:
$$\text{ (4) }k-2* \sum_{\substack{0\le i\lt P}}A_i=A_p$$

Now we can write the code to solve this problem, I will use javascript on this case:

function solution(A) {
var total=0;
var sum=0;
for(var index in A){
total+=A[index];
}
for(var index in A){
if(total-2*sum===A[index]){
return parseInt(index);
}
sum+=A[index];
}
return -1;
}

This was the result:

You can see the results on this link: https://codility.com/demo/results/demoX2XQGX-U7K/