Why did index go out of range - swift

Question

Hello i'm doing the simple FizzBuzz challenge. I figured out a solution to the problem. But I don't understand why my first attempt went out of range or what is happening behind the scenes.
To stop it from going out of range I changed
array.remove(at: i)
to
array.remove(at: i - 1)
Below is my entire project.
First attempt
var array = [String]()
for i in 1...100{
array.append("\(i)")
if i % 3 == 0 && i % 5 == 0 {
array.remove(at: i)
array.append("FizzBuzz")
}else if i % 3 == 0{
array.remove(at: i)
array.append("Fizz")
} else if i % 5 == 0{
array.remove(at: i)
array.append("Buzz")
}
}
print(array)
Second attempt
var array = [String]()
for i in 1...100{
array.append("\(i)")
if i % 3 == 0 && i % 5 == 0 {
array.remove(at: i - 1 )
array.append("FizzBuzz")
}else if i % 3 == 0{
array.remove(at: i - 1)
array.append("Fizz")
} else if i % 5 == 0{
array.remove(at: i - 1)
array.append("Buzz")
}
}
print(array)

Answer 1

Because arrays are zero index and you're starting at 1 in your for loop, so when you get to three and try to retrieve the third element (which indicates they're four elements in the array) swift will throw an error. The simple solution will be to start at 0
for i in 0...100{
print(i)
array.append("\(i)")
if i % 3 == 0 && i % 5 == 0 {
array.remove(at: i)
array.append("FizzBuzz")
}else if i % 3 == 0{
array.remove(at: i)
array.append("Fizz")
} else if i % 5 == 0{
array.remove(at: i)
array.append("Buzz")
}
}
print(array)
}

Answer 2

The remove(at:) function removes the element at the specified position. For example, if I have an array like [1,2,3,4] and I call array.remove(at: 2) the resulting array would be [1,2,4].
It is important to understand that the function removes the element at the specified INDEX location. Not the values itself. Swift arrays are zero indexed meaning, the first element is at location 0.
Now with this knowledge, I suggest you go ahead and debug your first code with some print statements and you be able to quickly understand why your code is going out of range. Hopefully through this experiment you can understand some basic concepts.
Good luck!

Answer 3

You are saying
for i in 1...100{
But array indices start at 0. So on every loop the item you just added has index i-1. There is no item with index i yet.
To put it another way, consider this part of your code:
if i % 3 == 0 && i % 5 == 0 {
array.remove(at: i)
Here you are prevaricating on the meaning of i. In the first line, it is what number i is. But in the second line, it is what index i is at. And because you started with 1, those two numbers are apart by 1. For example, the first time thru the loop, i is 1, but the value 1 is at index 0.