## Instructions

Given a triangle of consecutive odd numbers:

```
1
3 5
7 9 11
13 15 17 19
21 23 25 27 29
```

find the triangle's row knowing its index (the rows are 1-indexed), e.g.:

odd_row(1) == [1]

odd_row(2) == [3, 5]

odd_row(3) == [7, 9, 11]

Note: your code should be optimized to handle big inputs.

## My solution:

```
function oddRow(n) {
let first = 1;
for(let i = 1; i<n; i++){
first+=i*2
}
let r = [first];
for(let i = 1; i<n; i++){
r.push(first+=2)
}
return r
}
```

## Explanation

First I got the first value of the row that I'm searching on, I did this by using a for loop that will iterate the number of rows, and in every iteration I changed the value of the var "first" by adding 2 to it and revaluing it.

```
for(let i = 1; i<n; i++){
first+=i*2
}
```

After that with that value I made the variable "r" that will contain the result that will be returned, inside of it I stored the first number of the row

```
let r = [first];
```

After that I used another loop and inside of it, I pushed every value after the first plus 2, and at the end I just returned "r"

```
for(let i = 1; i<n; i++){
r.push(first+=2)
}
return r
```

**What do you think about this solution? ðŸ‘‡ðŸ¤”**

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