Matching strings, consider some characters are the same - objective-c

Question

please help me with this problem.
I want to check if the targetString match the keyword or not. Consider some character may different, but should still return true.
Example:
targetString = #"#ß<"
keyword = #"abc", #"∂B(", #"#Aß<"
result: all must return true.
(Matched.targetString and all keyword are the same.)
Consider me have an array, contains list of character set that can be the same:
NSArray *variants = [NSArray arrayWithObjects:#"aA#∂", #"bBß", #"c©C<(", nil]
So that when matching, with this rule, it can match as the example above.
Here is what i've done so far (using recursion):
- (BOOL) test:(NSString*)aString include:(NSString*) keyWord doTrim:(BOOL)doTrim {
// break recursion.
if([aString length] < [keyWord length]) return false;
// First, loop through each keyword's character
for (NSUInteger i = 0; i < [keyWord length]; i++) {
// Get #"aA#∂", #"bBß", #"c©C<(" or only the character itself.
// like, if the keyword's character is A, return the string #"aA#∂".
// If the character is not in the variants set, eg. P, return #"P"
char c = [keyWord characterAtIndex:i];
NSString *rs = [self variantsWithChar:c];
// Check if rs (#"aA#∂" or #"P") contains aString[i] character
if([rs rangeOfString:[NSString stringWithCharacters:[aString characterAtIndex:i] length:1]].location == NSNotFound) {
// If not the same char, remove first char in targetString (aString), recursion to match again.
return [self test:[aString substringFromIndex:1] include:keyWord doTrim:NO];
}
}
// If all match with keyword, return true.
return true;
}
- (NSString *) variantsWithChar:(char) c {
for (NSString *s in self.variants) {
if ([s rangeOfString:[NSString stringWithFormat:#"%c",c]].location != NSNotFound) {
return s;
}
}
return [NSString stringWithFormat:#"%c", c];
}
The main problem is, variantsWithChar: doesn't return the correct string. I don't know which datatype and which function should I use here. Please help.
For thou who know ruby, here's the example in ruby. It work super fine!
require 'test/unit/assertions'
include Test::Unit::Assertions
class String
def matching?(keyword)
length >= keyword.length && (keyword.chars.zip(chars).all? { |cs| variants(cs[0]).include?(cs[1]) } || slice(1, length - 1).matching?(keyword))
end
private
VARIANTS = ["aA#∂", "bBß", "c©C<("]
def variants(c)
VARIANTS.find { |cs| cs.include?(c) } || c
end
end
assert "abc".matching?("#ß<")
PS: The fact is, it's containt a japanese character set that sounds the same (like あア, いイ... for thou who know japanese)
PS 2: Please feel free to edit this Question, since my engrish is sooo bad. I may not tell all my thought.
PS 3: And, maybe some may comment about the performance. Like, search about 10,000 target words, with nearly 100 variants, each variant have at most 4 more same characters.

Answer 1

So first off, ignore comments about ASCII and stop using char. NSString and CFString use unichar
If what you really want to do is transpose hiragana and katakana you can do that with CFStringTransform()
It wraps the ICU libraries included in OS X and iOS.
It makes it very simple.
Search for that function and you will find examples of how to use it.

Answer 2

After a while (a day) working on the code above, I finally get it through. But don't know about the performance. Someone comment and help me improve about performance, please. Thanks.
- (BOOL) test:(NSString*)aString include:(NSString*) keyWord doTrim:(BOOL)doTrim {
// break recursion.
if([aString length] < [keyWord length]) return false;
// First, loop through each keyword's character
for (NSUInteger i = 0; i < [keyWord length]; i++) {
// Get #"aA#∂", #"bBß", #"c©C<(" or only the character itself.
// like, if the keyword's character is A, return the string #"aA#∂".
// If the character is not in the variants set, eg. P, return #"P"
NSString* c = [NSString stringWithFormat:#"%C", [keyWord characterAtIndex:i]];
NSString *rs = [self variantsWithChar:c];
NSString *theTargetChar = [NSString stringWithFormat:#"%C", [aString characterAtIndex:i]];
// Check if rs (#"aA#∂" or #"P") contains aString[i] character
if([rs rangeOfString:theTargetChar].location == NSNotFound) {
// If not the same char, remove first char in targetString (aString), recursion to match again.
return [self test:[aString substringFromIndex:1] include:keyWord doTrim:NO];
}
}
// If all match with keyword, return true.
return true;
}
If you remove all comment, it'll be pretty short...
////////////////////////////////////////
- (NSString *) variantsWithChar:(NSString *) c{
for (NSString *s in self.variants) {
if ([s rangeOfString:c].location != NSNotFound) {
return s;
}
}
return c;
}

Answer 3

You could try comparing ascii values of the japanese characters in the variants's each character's ascii value. These japanese characters aren't treated like usual characters or string. Hence, string functions like rangeOfString won't work on them.
to be more precise: have a look at the following code.
it will search for "∂" in the string "aA#∂"
NSString *string = #"aA#∂";
NSMutableSet *listOfAsciiValuesOfString = [self getListOfAsciiValuesForString:string]; //method definition given below
NSString *charToSearch = #"∂";
NSNumber *ascii = [NSNumber numberWithInt:[charToSearch characterAtIndex:0]];
int countBeforeAdding = [listOfAsciiValuesOfString count],countAfterAdding = 0;
[listOfAsciiValuesOfString addObject:ascii];
countAfterAdding = [listOfAsciiValuesOfString count];
if(countAfterAdding == countBeforeAdding){ //element found
NSLog(#"element exists"); //return string
}else{
NSLog(#"Doesnt exists"); //return char
}
===================================
-(NSMutableSet*)getListOfAsciiValuesForString:(NSString*)string{
NSMutableSet *set = [[NSMutableSet alloc] init];
for(int i=0;i<[string length];i++){
NSNumber *ascii = [NSNumber numberWithInt:[string characterAtIndex:i]];
[set addObject:ascii];
}
return set;
}