NSArray *a = [[NSArray alloc] initWithObjects:@(1),@(3),@(4),@(5),@(5), nil]; NSArray *b = [[NSArray alloc] initWithObjects:@(-1),@(3),@(0),@(9),@(4), nil]; NSArray *c = [[NSArray alloc] initWithObjects:@(0),@(31),@(32),@(22),@(6), nil]; NSOrderedSet *orderedSetA = [NSOrderedSet orderedSetWithArray:a]; NSOrderedSet *orderedSetB = [NSOrderedSet orderedSetWithArray:b]; NSOrderedSet *orderedSetC = [NSOrderedSet orderedSetWithArray:c]; NSMutableArray *collectionArray = [[NSMutableArray alloc] init]; NSMutableArray *resultArray = [[NSMutableArray alloc] init]; [collectionArray addObjectsFromArray:[orderedSetA array]]; [collectionArray addObjectsFromArray:[orderedSetB array]]; [collectionArray addObjectsFromArray:[orderedSetC array]]; NSCountedSet *cs = [[NSCountedSet alloc] initWithArray:collectionArray]; for (NSNumber *number in cs) { if ([cs countForObject:number] > 1) { if (![resultArray containsObject:number]) { [resultArray addObject:number]; } } } NSLog(@"%@",resultArray);

I didn't answered well. Totally dead, wrote tons and tons of code in xcode, but when it comes to simple questions like this and that too in a notepad without Xcode, my brain totally out of station. But yeah I tried something, I used to combine all three arrays into a mutable array and put an NSOrderedSet to remove the duplicates. Then run a loop with new array and find the range the range of objects presented in all three arrays. But they asked about O(n) complexity, time complexity for my solution, which I don't know totally. Update: I failed purely because I didn't had a Xcode and almost we can't memorise all the syntax and apis that iOS provides. After the interview I opened the xcode and found an abstract api in NSArray, to find common elements between 2 arrays, then I solved this problem within 2 minutes. Here is the answer below. Tips: Ask the interviewer whether you can use Xcode IDE for testing your codes. I didn't tried it. Answer: - (NSArray *)findAtleastTwoElementsPresentInTwoArray:(NSArray *)a bA:(NSArray *)b cA:(NSArray *)c{ NSMutableArray *elements = [NSMutableArray array]; id objcCommon1 = [a firstObjectCommonWithArray:b]; id objcCommon2 = [b firstObjectCommonWithArray:c]; id objcCommon3 = [c firstObjectCommonWithArray:a]; if (objcCommon1){ [elements addObject:objcCommon1]; } if (objcCommon2) { [elements addObject:objcCommon2]; } if (objcCommon3) { [elements addObject:objcCommon1]; } if (elements.count >= 2) { return elements; }else{ return nil; } }

+ (NSArray*) findCommonElements:(NSArray*)a and:(NSArray*)b and:(NSArray*)c { NSMutableArray * common = [NSMutableArray new]; NSMutableArray * combined = [NSMutableArray new]; [combined addObjectsFromArray:a]; [combined addObjectsFromArray:b]; [combined addObjectsFromArray:c]; NSCountedSet * set = [[NSCountedSet alloc]initWithArray:combined]; for (id obj in set) { if ([set countForObject:obj] >= 2) { [common addObject:obj]; } } return common; }

union(intersection (set(array1), set(array2)) , intersection (set(array2), set(array3)) , intersection (set(array1), set(array3)))

Put elements all three array in 3 sets, then put all the elements from these sets to a counted set (NSCountedSet), now just use a predicate to fined which all elements are there in this final set with count more than 2.

NSArray *a = @[@1, @2, @3]; NSArray *b = @[@3, @4, @5]; NSArray *c = @[@5, @6, @7]; NSMutableSet *aSet = [NSMutableSet setWithArray:a]; NSMutableSet *bSet = [NSMutableSet setWithArray:b]; NSSet *cSet = [NSSet setWithArray:c]; [aSet intersectSet:bSet]; [bSet intersectSet:cSet]; [aSet unionSet:bSet]; NSLog(@"Final elements are:%@", aSet);