Edited: Below is an interview coding test:
In my opinion this should be a staff training exercise instead on an interview test.

This is a demo task.

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[]);

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.

Source: https://codility.com/demo/take-sample-test/

It is kind of simple, the only problem is just the required O(N);

^^^
The coding and O(N) is no big deal for a programmer with CS of SE.

como se dice "CS del SE" en ingles?

^^^
Na so e be o!

How do you say "CS of SE" in English?

^^^
Now moving from programming to Pro-Grammar

HAHAHAHA.

No comprende?

dhtml18:
No comprende?

Did u dream of france last night

France? uhm you are mistaken mi amigo, estoy spanish.

I will break down the 1st part of the task just to make it simpler for those that might like to attempt it.

A[0] + A[1] + ... + A[P−1] = A[P+1] + ... + A[N−2] + A[N−1]

Can be interpreted as: A[0] + A[1] + A[3-1] = A[3+1] + A[8−2] + A[8−1]

Where:
N = 3 which is the length of the array
and
P = 3

dhtml18:
France? uhm you are mistaken mi amigo, estoy spanish.

OOPS! i thought it was french.

Hehehehehehe, with some certain constructs, it is a bit hard to figure out whether it is french or spanish.

ne comprendre (french) === no comprende (spanish) === not understand (english)

I can only speak/write spanish to an extent, but my french is quite quite poor like an average Nigerian.

Lets come back to the quiz, it looks like Febup is trying to win o, crotonite show him that you are a true programmer jor!

Currently busy at the moment @dhtml18. let him win, i will hustle 4 2nd or 3rd place.

Chai, me too busy small - battling a native android app that streams live videos from a server - the thing don nearly finish me for here!

^^^
Okay-dokay, after you finish the almighty android app you can come back and take the quiz.

dhtml18:
Chai, me too busy small - battling a native android app that streams live videos from a server - the thing don nearly finish me for here!

native as in c/c++

dhtml18:
Chai, me too busy small - battling a native android app that streams live videos from a server - the thing don nearly finish me for here!

what language boss I know you will soon build the app . hearing server I guess u are using html5 .

crotonite:
native as in c/c++

JAVA/Android Studio

Here...Just got this off the top of my head. I know it has flaws but till I get a better solution and I will:

public static int solution(int[] nums, int n){

if(sum(nums)==0) return 0;

int index = 0;
int i = 1;


while( i < nums.length){

int[] left = Arrays.copyOfRange(nums, 0, i);
int[] right = Arrays.copyOfRange(nums, i+1, nums.length);

//1, 2, 3, 5, 1, 2, 3, 0

if( sum(left)== sum(right)){
return i;
}
i++;
}

return -1;

}

private static int sum(int[] arr){

int a = 0;
for(int i: arr){
a+=i;
}
return a;
}

^^^O boy, see code o - this is the kind of codes that makes me start to dance alanta!

larisoft:

Here...Just got this off the top of my head. I know it has flaws but till I get a better solution and I will:

public static int solution(int[] nums, int n){

if(sum(nums)==0) return 0;

int index = 0;
int i = 1;


while( i < nums.length){

int[] left = Arrays.copyOfRange(nums, 0, i);
int[] right = Arrays.copyOfRange(nums, i+1, nums.length);

//1, 2, 3, 5, 1, 2, 3, 0

if( sum(left)== sum(right)){
return i;
}
i++;
}

return -1;

}

private static int sum(int[] arr){

int a = 0;
for(int i: arr){
a+=i;
}
return a;
}

This runs in O(n).

public static int solution2(int[] nums, int n){

int sum = sum(nums);

int i = 0;
int left_sum = 0;

while(i < nums.length){

int right_sum = sum - left_sum - nums[i];

if(right_sum == left_sum){
return i;
}

left_sum+= nums[i];
i++;

}

return -1;

}

private static int sum(int[] arr){

int a = 0;
for(int i: arr){
a+=i;
}
return a;
}

@larisoft
You are the LAW on the this programming section

^^^^ #fact, especially now wey i don relegate for top 500 programmers on nairaland chart (last i checked).

^^^
You've been too busy with the almighty apps you are working on, but don't worry as your are still the OGA on this section.

Febup:
@larisoft You are the LAW on the this programming section

lol...thanks, bro.

just trollimg

www.nairaland.com/attachments/3721007_20160516151740_jpeg513e84d55984691f1325a165ce18ac39

Anoda troll on this section?

Febup:
@larisoft
You are the LAW on the this programming section

my oga dhtml18 transcends him....

crotonite:
my oga dhtml18 transcends him....

Pin.chur or adonbilivit

dhtml18:

Pinchur or adonbilivit

lol