How to compare two NSArrays of NSStrings, ignoing case and order

Question

I have two NSArrays, each containing NSStrings.

How do I check they are equivalent, ignoring case and order?

I've been playing around with NSSets and NSPredicates, but not having much luck.

Answer 1

Simple (case insensitive)

NSSet* one = [[NSCountedSet setWithArray:firstArray] valueForKey:@"lowercaseString"];
NSSet* two = [[NSCountedSet setWithArray:secondArray] valueForKey:@"lowercaseString"];

return [one isEqualToSet:two];

---

Complex (case and diacritic insensitive)

This solution is more complex, because it requires a special string comparison. You can either accomplish this by iterating sorted arrays using this method, or, if you wish to use valueForKey:, using the following code.

@interface LNComparisonHelper : NSObject

- (instancetype)initWithString:(NSString*)str;
@property (nonatomic, copy, readonly) string;

@end

@implementation LNComparisonHelper

- (instancetype)initWithString:(NSString*)str
{
    self = [super init];
    if(self) { _str = [str copy]; }
    return self;
}

- (BOOL)isEqual:(LNComparisonHelper*)other
{
    return [_str compare:other.string options:(DiacriticInsensitiveSearch | CaseInsensitiveSearch)] == OrderedSame;
}

@end

@interface NSString (ComparisonWrapper)

- (id) ln_casediacriticInsensitiveComparator;

@end

@implementation NSString (ComparisonWrapper)

- (id) ln_casediacriticInsensitiveComparator
{
    return [[LNComparisonHelper alloc] initWithString:self];
}

@end

NSSet* one = [[NSCountedSet setWithArray:firstArray] valueForKey:@"ln_casediacriticInsensitiveComparator"];
NSSet* two = [[NSCountedSet setWithArray:secondArray] valueForKey:@"ln_casediacriticInsensitiveComparator"];

return [one isEqualToSet:two];

Answer 2

1. Compare array counts
2. Copy into new NSMutableArrays
3. Sort the arrays (sortUsingComparator: with NSCaseInsensitiveSearch and DiacriticInsensitiveSearch)
4. By pairs in order compare strings (NSCaseInsensitiveSearch and DiacriticInsensitiveSearch)

As soon as there is a failure return NO
If no failure YES

Note: For reasonably sized arrays sorting has a low cost close to O(n).