Algorithm to find anagrams Objective-C - objective-c

Question

I've got an algorithm to find anagrams within a group of eight-letter words. Effectively it's alphabetizing the letters in the longer word, doing the same with the shorter words one by one, and seeing if they exist in the longer word, like so:
tower = eortw
two = otw
rot = ort
The issue here is that if I look for ort in eortw (or rot in tower), it'll find it, no problem. Rot is found inside tower. However, otw is not inside eortw (or two in tower), because of the R in the middle. Ergo, it doesn't think two is found in tower.
Is there a better way I can do this? I'm trying to do it in Objective-C, and both the eight-letter words and regular words are stored in NSDictionaries (with their normal and alphabetized forms).
I've looked at various other posts re. anagrams on StackOverflow, but none seem to address this particular issue.
Here's what I have so far:
- (BOOL) doesEightLetterWord: (NSString* )haystack containWord: (NSString *)needle {
for (int i = 0; i < [needle length] + 1; i++) {
if (!needle) {
NSLog(#"DONE!");
}
NSString *currentCharacter = [needle substringWithRange:NSMakeRange(i, 1)];
NSCharacterSet *set = [NSCharacterSet characterSetWithCharactersInString: currentCharacter];
NSLog(#"Current character is %#", currentCharacter);
if ([haystack rangeOfCharacterFromSet:set].location == NSNotFound) {
NSLog(#"The letter %# isn't found in the word %#", currentCharacter, haystack);
return FALSE;
} else {
NSLog(#"The letter %# is found in the word %#", currentCharacter, haystack);
int currentLocation = [haystack rangeOfCharacterFromSet: set].location;
currentLocation++;
NSString *newHaystack = [haystack substringFromIndex: currentLocation];
NSString *newNeedle = [needle substringFromIndex: i + 1];
NSLog(#"newHaystack is %#", newHaystack);
NSLog(#"newNeedle is %#", newNeedle);
}
}
}

Answer 1

If you use only part of the letters it isn't a true anagram.
A good algorithm in your case would be to take the sorted strings and compare them letter by letter, skipping mis-matches in the longer word. If you reach the end of the shorter word then you have a match:
char *p1 = shorter_word;
char *p2 = longer_word;
int match = TRUE;
for (;*p1; p1++) {
while (*p2 && (*p2 != *p1)) {
p2++;
}
if (!*p2) {
/* Letters of shorter word are not contained in longer word */
match = FALSE;
}
}

Answer 2

This is one that approach I might take for finding out if one ordered word contained all of the letters of another ordered word. Note that it won't find true anagrams (That simply requires the two ordered strings to be the same) but this does what I think you're asking for:
+(BOOL) does: (NSString* )longWord contain: (NSString *)shortWord {
NSString *haystack = [longWord copy];
NSString *needle = [shortWord copy];
while([haystack length] > 0 && [needle length] > 0) {
NSCharacterSet *set = [NSCharacterSet characterSetWithCharactersInString: [needle substringToIndex:1]];
if ([haystack rangeOfCharacterFromSet:set].location == NSNotFound) {
return NO;
}
haystack = [haystack substringFromIndex: [haystack rangeOfCharacterFromSet: set].location+1];
needle = [needle substringFromIndex: 1];
}
return YES;
}

Answer 3

The simplest (but not most efficient) way might be to use NSCountedSet. We can do this because for counted sets, [a isSubsetOfSet:b] return YES if and only if [a countForObject:object] <= [b countForObject:object] for every object in a.
Let's add a category to NSString to do it:
#interface NSString (lukech_superset)
- (BOOL)lukech_isSupersetOfString:(NSString *)needle;
#end
#implementation NSString (lukech_superset)
- (NSCountedSet *)lukech_countedSetOfCharacters {
NSCountedSet *set = [NSCountedSet set];
[self enumerateSubstringsInRange:NSMakeRange(0, self.length) options:NSStringEnumerationByComposedCharacterSequences usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
[set addObject:substring];
}];
return set;
}
- (BOOL)lukech_isSupersetOfString:(NSString *)needle {
return [[needle lukech_countedSetOfCharacters] isSubsetOfSet:[self lukech_countedSetOfCharacters]];
}
#end