Question

I intend to make a program that does the following:

Create an NSArray populated with numbers from 1 to 100,000.
Loop over some code that deletes certain elements of the NSArray when certain conditions are met.
Store the resultant NSArray.

However the above steps will also be looped over many times and so I need a fast way of making this NSArray that has 100,000 number elements.

So what is the fastest way of doing it?

Is there an alternative to iteratively populating an Array using a for loop? Such as an NSArray method that could do this quickly for me?

Or perhaps I could make the NSArray with the 100,000 numbers by any means the first time. And then create every new NSArray (for step 1) by using method arraywithArray? (is it quicker way of doing it?)

Or perhaps you have something completely different in mind that will achieve what I want.

edit: replace NSArray with NSMutableArray in above post

Answer 1

You may want to look at NSMutableIndexSet. It is designed to efficiently store ranges of numbers.

You can initialize it like this:

NSMutableIndexSet *set = [[NSMutableIndexSet alloc]
    initWithIndexesInRange:NSMakeRange(1, 100000)];

Then you can remove, for example, 123 from it like this:

[set removeIndex:123];

Or you can remove 400 through 409 like this:

[set removeIndexesInRange:NSMakeRange(400, 10)];

You can iterate through all of the remaining indexes in the set like this:

[set enumerateIndexesUsingBlock:^(NSUInteger i, BOOL *stop) {
    NSLog(@"set still includes %lu", (unsigned long)i);
}];

or, more efficiently, like this:

[set enumerateRangesUsingBlock:^(NSRange range, BOOL *stop) {
    NSLog(@"set still includes %lu indexes starting at %lu",
        (unsigned long)range.length, (unsigned long)range.location);
}];