Prepping for an interview. I am trying to practice by solving the following problem: Given an input array of NSNumbers where some of the numbers are duplicated, how can you create another array that only has the unique values in the original array.
I see 2 approaches:
Brute-force: Loop through each element in the array, while at a element compare it against the set of numbers in the unique list, if there is a match, don't store it, else add it to the unique list. O(n^2) worst case time?
Hash-table based approach: Have a hash-table of length N. Each element of the has-table is NSSet. Every number is mapped to 0,...N-1 using a hashing function. If it is exists in the NSSet corresponding to the "mapped-index", it is not added to "unique array". if not, it is added to set and unique array.
Is this O(N) complexity?
Code for each approach is below.
I am noticing that
i. Running time of 2A (array approach) is half of that of 2B (Mutabledictionary approach) for the input array of length 403 shown below(0.055ms vs .12ms).
ii. Running time of 1 is ~ 5 times worse 0.25ms. If there are not any duplicates, this discrepancy is even worse.
My Qs are:
Code
Hashcode function
#define NUM_BUCKETS 127
#define RANDOMIZER 11
#define NUM_ITER 40000
int hashcode(int value)
{
int retVal = (value*RANDOMIZER)%NUM_BUCKETS ;
if(retVal<0)
{
retVal+=NUM_BUCKETS ;
}
return retVal ;
}
1. Brute-Force Approach
NSMutableArray *smooshedArr=[[NSMutableArray alloc] init] ;
double startTime ;
startTime=CFAbsoluteTimeGetCurrent() ;
for(int iter=0;iter<=NUM_ITER;iter++)
{
[smooshedArr removeAllObjects] ;
[smooshedArr addObject:ints[0]] ;
int i,j ;
for(i=1;i<[ints count];i++)
{
for(j=0;j<[smooshedArr count];j++)
{
if([ints[i] intValue] == [smooshedArr[j] intValue])
{
break ;
}
}
if(j==[smooshedArr count])
{
[smooshedArr addObject:ints[i]] ;
}
}
}
NSLog(@"Bruteforce took %.3fms to remove duplicates from array of length %lu",(CFAbsoluteTimeGetCurrent()-startTime)*1000/NUM_ITER,(unsigned long)[ints count]) ;
NSLog(@"Smooshed arary is %@",smooshedArr) ;
2A. Array based hash table
NSMutableArray *hashTable = [[NSMutableArray alloc] init] ;
startTime=CFAbsoluteTimeGetCurrent() ;
for(int iter=0;iter<=NUM_ITER;iter++)
{
[smooshedArr removeAllObjects];
for (NSInteger i = 0; i < NUM_BUCKETS; ++i)
{
[hashTable addObject:[NSNull null]];
}
[smooshedArr addObject:ints[0]] ;
int indexToInsert = hashcode([ints[0] intValue]) ;
hashTable[indexToInsert]=[[NSMutableSet alloc] init] ;
[hashTable[indexToInsert] addObject:ints[0]] ;
int i ;
for(i=1;i<[ints count];i++)
{
//Find hascode of element i
//If the list at index = hashcode in hashCodeArary is empty, then create a NSMutableSet, set toInsert = True
//If not empty, check if the element exists in the set. If yes, setToInsert=False. If no, setToInsert=True
int indexToInsert = hashcode([ints[i] intValue]) ;
BOOL toInsert=false ;
if(hashTable[indexToInsert] == [NSNull null])
{
hashTable[indexToInsert]=[[NSMutableSet alloc] init] ;
toInsert=true ;
}
else
{
if(![hashTable[indexToInsert] containsObject:ints[i]])
toInsert=true ;
}
if(toInsert)
{
[hashTable[indexToInsert] addObject:ints[i]] ;
[smooshedArr addObject:ints[i]] ;
}
}
}
NSLog(@"MutableArray (no cheat) took %.3fms to remove duplicates from array of length %lu",(CFAbsoluteTimeGetCurrent()-startTime)*1000/NUM_ITER,(unsigned long)[ints count]) ;
2B. Dictionary based hash table
NSMutableDictionary *hashDict = [[NSMutableDictionary alloc] init] ;
//NSLog(@"Start of hashcode approach %.6f", CFAbsoluteTimeGetCurrent()) ;
startTime=CFAbsoluteTimeGetCurrent() ;
for(int iter=0;iter<=NUM_ITER;iter++)
{
//if(iter <4) NSLog(@"iter start: %.6f", CFAbsoluteTimeGetCurrent()) ;
//if(iter <4) NSLog(@"init start: %.6f", CFAbsoluteTimeGetCurrent()) ;
[smooshedArr removeAllObjects];
[hashDict removeAllObjects] ;
//if (iter<4) NSLog(@"init end: %.6f", CFAbsoluteTimeGetCurrent()) ;
[smooshedArr addObject:ints[0]] ;
int indexToInsert = hashcode([ints[0] intValue]) ;
hashDict[@(indexToInsert)]=[[NSMutableSet alloc] init] ;
[hashDict[@(indexToInsert)] addObject:ints[0]] ;
int i ;
for(i=1;i<[ints count];i++)
{
//Find hascode of element i
//If the list at index = hashcode in hashCodeArary is empty, then create a NSMutableSet, set toInsert = True
//If not empty, check if the element exists in the set. If yes, setToInsert=False. If no, setToInsert=True
int indexToInsert = hashcode([ints[i] intValue]) ;
BOOL toInsert=false ;
if(hashDict[@(indexToInsert)] == nil)
{
hashDict[@(indexToInsert)]=[[NSMutableSet alloc] init] ;
toInsert=true ;
}
else
{
if(![hashDict[@(indexToInsert)] containsObject:ints[i]])
toInsert=true ;
}
if(toInsert)
{
[hashDict[@(indexToInsert)] addObject:ints[i]] ;
[smooshedArr addObject:ints[i]] ;
}
}
}
NSLog(@"Dictionary approach: %.3fms to remove duplicates from array of length %lu",(CFAbsoluteTimeGetCurrent()-startTime)*1000/NUM_ITER,(unsigned long)[ints count]) ;
Input tested ON, 430 elements with some dups and averaged over 40000 iterations
NSArray *ints = @[@(2),@(3),@(4),@(1),@(6),@(9),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(65),@(45454),@(1234),@(3421),@(123),@(1234),@(11111),@(2727272),@(112),@(3),@(4),@(1),@(612211),@(9),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(65),@(45454),@(1234),@(3421),@(123),@(1234),@(11111),@(7272),@(1232),@(3),@(4),@(1),@(60),@(9),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(65),@(45454),@(1234),@(3421),@(123),@(1234),@(11111),@(2727272),@(2),@(3),@(4),@(1),@(6),@(9),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(65),@(45454),@(1234),@(3421),@(123),@(1234),@(11111),@(2727272),@(2),@(3),@(4),@(1),@(6),@(9),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(65),@(45454),@(1234),@(3421),@(123),@(1234),@(11111),@(72),@(2),@(3),@(4),@(1),@(6),@(9),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(65),@(45454),@(1234),@(3421),@(123),@(1234),@(11111),@(13272),@(2),@(3),@(4),@(18),@(6),@(9),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(65),@(45454),@(1234),@(3421),@(123),@(1234),@(11111),@(972),@(2),@(3),@(4),@(1),@(6),@(9),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(65),@(45454),@(1234),@(3421),@(123),@(1234),@(11111),@(3272),@(2),@(3),@(4),@(1),@(69),@(9),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(65),@(45454),@(1234),@(3421),@(123),@(1234),@(11111),@(1272),@(2),@(3),@(4),@(1),@(6),@(9),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(65),@(45454),@(1234),@(3421),@(123),@(1234),@(11111),@(2272),@(2),@(3),@(4),@(1),@(6),@(91),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(65),@(45454),@(1234),@(3421),@(123),@(1234),@(11111),@(7272),@(2),@(3),@(4),@(12),@(6),@(9),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(65),@(45454),@(1234),@(3421),@(123),@(1234),@(111),@(27272),@(2),@(321),@(4),@(1),@(6),@(9),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(65),@(4545411),@(12341),@(34210),@(123),@(1234),@(1111),@(727272),@(11187),@(9086),@(876543),@(74532),@(464642),@(65),@(45454),@(1234),@(3421),@(123),@(1234),@(11111),@(13272),@(2),@(3),@(4),@(18),@(6),@(9),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(658),@(45454),@(12934),@(38421),@(1243),@(12345),@(1112),@(72),@(52),@(3),@(498),@(1),@(6),@(9),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(65),@(45454),@(1234),@(650),@(45454),@(1234),@(3421),@(123),@(1234),@(111),@(27272),@(2),@(321),@(4),@(1),@(6),@(9),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(65),@(4545411),@(12341),@(34210),@(123),@(1234),@(1111),@(727272),@(11187),@(9086),@(876543),@(74532),@(464642),@(65),@(45454),@(1234),@(3421),@(123),@(1234),@(11111),@(13272),@(2),@(3),@(4),@(18),@(6),@(9),@(2),@(2),@(3),@(21),@(22),@(45454),@(464642),@(658),@(45454),@(12934),@(38421),@(1243),@(19992345),@(119875412),@(72),@(52),@(3),@(498),@(1),@(6),@(9),@(2),@(2),@(3),@(21),@(22),@(450454),@(46908764642),@(6753435),@(45498754),@(100234),@(65)] ;
If you are preparing for an interview, I would advise you to use the framework classes that are already implemented. Don't reimplement the wheel. Try to solve the problem from top to bottom. Don't think about details (hash functions), think about the algorithm structure:
In pseudocode:
for number in input {
if number appears for the first time {
add number to output
}
}
The only problem we have is how to implement the number appears for the first time. That's the only point that has some performance implications here.
In Objective-C we can use NSSet which is a class created exactly for this problem.
NSArray *input = @[... array of numbers];
NSMutableSet *foundNumbers = [NSMutableSet set];
NSMutableArray *output = [NSMutableArray array];
for (NSNumber *number in input) {
if (![foundNumbers containsObject:number])) {
[foundNumbers addObject:number];
[output addObject:number];
}
}
NSLog(@"Output: %@", output);
You need only one pass of the input array. The only way you could improve performance is using a different structure than NSSet, however NSSet is already highly optimized and it's unlikely you will find a better option.
If you want to think out of the box and the numbers in your input are limited to a small enough range (e.g. 0...65000), you can create a BOOL array with 65000 items, all initialized to NO and use that as a fast set implementation.
However, that will take a lot of memory and it won't pay off unless the input array is very long.
Definitely don't implement your own hash tables, NSDictionary is already a hash table. What you are doing in your second implementation is just a very obfuscated reimplementation of NSDictionary. Buckets work only when you can keep them as a simple array. Once you add hash function to it, you are losing the performance gain.
Also note that the overall quality of code is very important for interviews. Don't use #define to declare a constant. Keep a good coding style (I would strongly advice to use spaces around operators). Use iterators instead of for(;;) Try to name your variables better than hashDict (name your variables for the data they contain).
Now a little secret, there is also a class NSOrderedSet which combines NSArray and NSSet into one object and can solve your problem even easier:
NSOrderedSet *orderedSet = [NSOrderedSet orderedSetWithArray:ints];
NSLog(@"Output: %@", orderedSet);