Daily Coding Challenge 56

DCC_56.m

/*
Given a dictionary of words and a string made up of those words (no spaces), return the original sentence in a list. If there is more than one possible reconstruction, return any of them. If there is no possible reconstruction, then return null.

For example, given the set of words 'quick', 'brown', 'the', 'fox', and the string "thequickbrownfox", you should return ['the', 'quick', 'brown', 'fox'].

Given the set of words 'bed', 'bath', 'bedbath', 'and', 'beyond', and the string "bedbathandbeyond", return either ['bed', 'bath', 'and', 'beyond] or ['bedbath', 'and', 'beyond'].

store words in a set
loop through string
	build word
	if word is in set
    add word to result
		word = blank

retrun result

O(n) time where n is the length of the string
O(n) space for storing the temporary word, also O(n) if we have to build the set or O(1) if given it
*/

-(NSArray<NSString*> *)splitWords:(NSString *)s setOfWords:(NSArray<NSString*> *)words {
  NSSet *set = [[NSSet alloc]initWithArray:words];
  NSMutableString *word = [[NSMutableString alloc] init];
  NSMutableArray *result = [[NSMutableArray alloc] init];
  for (int i = 0; i < s.length; ++i) {
    char c = [s characterAtIndex:i];
    [word appendString:@(c)];
    if ([set containsObject:word]) {
      [result addObject:word];
      word = @""
    }
  }
  if (result.count == 0)
    return nil
  return result
}