Get common character(s) from NSString as substring

Question

I'm trying to get repeated characters in NSString with their count.

For example, setupsetpreset as an input and it returns 'set' along with the count 3.

Let me explain with another example.

Input 1 : upsetpreset

Expected output 1: set 2

Input 2 : qwertygdgpopgdg

Expected output 2: gdg 2

obviously 2 and using that, I need to get occurrence of same phrase — Pradheep Narendran P, Mar 09 '16 at 08:22
Suppose the following string. Sheraheshehehesheshe What should be the output? As you can see "She" and "he" are BOTH repeated. — Ruchira Randana, Mar 09 '16 at 08:47
I agree. But, It will return the count of 'he' since its count is greater than 'she'. — Pradheep Narendran P, Mar 09 '16 at 08:54

Awesome.Apple · Answer 1 · 2016-03-09T08:38:18.397

2

Refer below

NSString *string     = @"This is a test. This is only a test.";
NSCountedSet *countedSet = [NSCountedSet new];

[string enumerateSubstringsInRange:NSMakeRange(0, [string length])
                       options:NSStringEnumerationByWords | NSStringEnumerationLocalized
                    usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop){

                        // This block is called once for each word in the string.
                        [countedSet addObject:substring];

                        // If you want to ignore case, so that "this" and "This" 
                        // are counted the same, use this line instead to convert
                        // each word to lowercase first:
                        // [countedSet addObject:[substring lowercaseString]];
                    }];

NSLog(@"%@", countedSet);

Results:

"This" - 2 "is" - 2 "a" - 2 "test" - 2 "only" - 1

edited Mar 09 '16 at 08:38

answered Mar 09 '16 at 08:22

Awesome.Apple

1,316
1
11
24

Thanks... But the term should be not mentioned along with the input. For example, input is setupsetpreset and the output returns a phrase 'set' – Pradheep Narendran P Mar 09 '16 at 08:23
you told you want only count. And it will return you the count. – Awesome.Apple Mar 09 '16 at 08:24
I mentioned an output returned with count – Pradheep Narendran P Mar 09 '16 at 08:25
But repeated character you already know will be @"set" if (count is > 1) – Awesome.Apple Mar 09 '16 at 08:28
Please forget about 'set'. Let's take another example of qwertysdspopsds. In this, sds is a phrase which is repeated. So, the expected output is sds 2. – Pradheep Narendran P Mar 09 '16 at 08:30
1

Refer this if it will work :- http://stackoverflow.com/questions/13366234/ios-most-efficient-way-to-find-word-occurrence-count-in-a-string – Awesome.Apple Mar 09 '16 at 08:34
Thanks. This is exactly what I'm looking for. – Pradheep Narendran P Mar 09 '16 at 08:37
I tried the link you sent. But, it returns the count of a character. I've added to a temporary array and still, how do I know whether a letter composed is subtring ? For example, deedddeede. In this example, de is also considered as substring and ed is also considered as substring since they're same in length.. – Pradheep Narendran P Mar 09 '16 at 08:57

Pradheep Narendran P · Answer 2 · 2016-03-09T11:03:37.610

After getting lot of custom logics to get common characters in a non-space string, I got the solution :-

NSString *string=@"setupsetpreset";
string=@"sheraheshehehesheshe";
NSMutableArray *a=[[NSMutableArray alloc] init];
NSMutableArray *b=[[NSMutableArray alloc] init];
for (int i=0; i<(string.length); i++) {
    [a addObject:[NSString stringWithFormat:@"%c",[string characterAtIndex:i]]];
}
NSCountedSet *countedSet = [NSCountedSet new];
[string enumerateSubstringsInRange:NSMakeRange(0, [string length])
                           options:NSStringEnumerationByComposedCharacterSequences
                        usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop){
                            [countedSet addObject:substring];
                        }];
NSMutableArray *c=[[NSMutableArray alloc] init];
for (int i=0; i<(string.length); i++) {
    if([countedSet countForObject:[a objectAtIndex:i]]>=2){
        if(![b containsObject:[a objectAtIndex:i]]){
            [c addObject:[NSString stringWithFormat:@"%d",(int)[countedSet countForObject:[a objectAtIndex:i]]]];
        }
        [b addObject:[a objectAtIndex:i]];
    }
}

int total=0;
for (NSString *s in c){
    int m=[s intValue];
    total+=m;
}
total=total/c.count;
NSString *s=[b componentsJoinedByString:@""];
[c removeAllObjects];
for (int i=0; i<(s.length); i++) {
    if([countedSet countForObject:[b objectAtIndex:i]]>=total){

        [c addObject:[b objectAtIndex:i]];
    }
}
s=[c componentsJoinedByString:@""];
int iterate=0;
NSString  *substring;
NSMutableDictionary *dict=[[NSMutableDictionary alloc] init];
while (iterate<s.length-total) {
    NSString *searchstring=[s substringWithRange:NSMakeRange(iterate, total)];
    if ([string rangeOfString:searchstring].location == NSNotFound) {
    } else {
            [dict setObject:[NSNumber numberWithInt:[[dict objectForKey:searchstring] intValue]+1] forKey:searchstring];
        }
        iterate++;
    }
    [a removeAllObjects];
    [b removeAllObjects];
    a=[dict.allValues mutableCopy];
    b=[dict.allKeys mutableCopy];
    int max = [[a valueForKeyPath:@"@max.intValue"] intValue];
    NSInteger Aindex = [a indexOfObject:[NSNumber numberWithInt:max]];
    substring =[NSString stringWithFormat:@"%@",[b objectAtIndex:Aindex]];
    NSLog(@"Substring %@ is repeated %d times",substring,max);

Hope it helps for someone's need :) Thanks Kiran for providing me a set of logic which is required as a part and Ruchira for your query.

I'm storing numbers inside an array to get the same index of the NSArray that is considered as substring matched in which the count is maximum. — Pradheep Narendran P, Mar 09 '16 at 10:57
Of course, it is possible using NSNumber. I've used it for maintaining a consistency with the data which is convenient for me. — Pradheep Narendran P, Mar 09 '16 at 10:59

Get common character(s) from NSString as substring

2 Answers2