You have this list

```
[7, 11, 0, 5, 6, 7]
```

And you iterate through its indices.

but you donâ€™t look at the current one, you look at the previous one, so youâ€™re really looking at this list:

```
[7, 7, 11, 0, 5, 6] # move last element to beginning
```

There canâ€™t possibly be any point to that

I donâ€™t see why you print the letter N, wouldnâ€™t it be better to know which value N has?

If youâ€™re trying to compare values pair-wise with the one that follows, why donâ€™t you start with iterating through the following:

```
(7, 11)
(11, 0)
(0, 5)
(5, 6)
(6, 7)
```

Note that this is one pass over the list, you would not be nesting loops inside each other

@stetim94

https://www.codewars.com/kata/equal-sides-of-an-array

How does comparing elements to each other relate to the problem?

Again with the example input of all values being equal - this wonâ€™t tell you anything

```
[1, 1, 1, 1, 1] # this has a solution: 2
[1, 1, 1, 1] # this has no solution (-1)
```

The problem statement is about *sums* there is a condition to check at each position:

find an index N where

the sum of the integers to the left of N is equal to the sum of the integers to the right of N

So for each possible N, you should be testing whether:

the sum of the integers to the left of N

is equal to

the sum of the integers to the right of N