Quote:Given 3 int values, a b c, return their sum. However, if any of the values is a teen -- in the range 13..19 inclusive -- then that value counts as 0, except 15 and 16 do not count as a teens. Write a separate helper "public int fixTeen(int n) {"that takes in an int value and returns that value fixed for the teen rule. In this way, you avoid repeating the teen code 3 times (i.e. "decomposition"). Define the helper below and at the same indent level as the main noTeenSum().

noTeenSum(1, 2, 3) → 6
noTeenSum(2, 13, 1) → 3
noTeenSum(2, 1, 14) → 3

Fiel Wrote:In this problem, there are four possible pairings which would cause exceptions. Either the first and second number, second and third, or third and first are alike. Each of the following if statements checks if the others are pairs.

 Spoiler

Code:

public int loneSum(int a, int b, int c) {
  int sum = 0;

  if(a^b > 0 && a^c > 0)
  {
      sum += a;
  }
  if(b^a > 0 && b^c > 0)
  {
      sum += b;
  }
  if(c^a > 0 && c^b > 0)
  {
      sum += c;
  }
  return sum;
}

I'm sure there must be a faster way to do this as well because b^a is the same as a^b.

Kalovale Wrote:Okay this is getting pretty petty of me, but I keep getting the impression that my approach isn't proper enough. It just seems somehow messy.

 Spoiler

I have a favor though, regarding this problem:

The solution should be simple enough, but I can't seem to "call" the n from the fixTeen block to the main block to use. Also, I may not be thinking straight but how do I filter so that the outcome n satisfies certain conditions? I have been doing something along the lines of conditionals:

Code:

public int fixTeen(int n){
  if( (n>=13) && (n<=19) && (n != 15) && (n != 16) )
  return n;
}

But I can't test if it works correctly because of the problem above.

Devil's Sunrise Wrote:As of right now, fixTeen returns n if n is 13, 14, 17, 18 or 19. Otherwise, what does it return? What should it return?

Kalovale Wrote:I expect it to return exactly those values, as for the "else" part, that's what perplexes me. I want to filter out the non-teen values with a view that after the method, n consists of only teens.

Devil's Sunrise Wrote:Wouldn't it be smarter for fixTeen to return 0 if it is 13, 14, 17, 18 or 19, and n otherwise? Then you can just do fixTeen(a) + fixTeen(b) + fixTeen©;

Quote:if (closeAB)

Kalovale Wrote:Same concept, but thank you. I know how to call n now.
I'm soooooooo abusing this, it makes things so much clearer.

Also, as for 4/, don't you mean for closeAB to be boolean? I mean, it's even used as a boolean. o.o

Kalovale Wrote:Also, is it noticeably bad practice to create unnecessary objects? Like the remainder below with is only used once and can easily be substituted by goal%5?

Kalovale Wrote:I feel more accomplished with this than with #4, but aside from sorting out all the special cases one by one, I can't think of another approach. Any suggestions/evaluations? I don't mean to be studying things I'm not supposed to. I'm just thinking there might be a fundamental technique that I missed from reading the guides.

Also, is it noticeably bad practice to create unnecessary objects? Like the remainder below with is only used once and can easily be substituted by goal%5?

 Spoiler

[COLOR="red"]We want make a package of goal kilos of chocolate. We have small bars (1 kilo each) and big bars (5 kilos each). Return the number of small bars to use, assuming we always use big bars before small bars. Return -1 if it can't be done.

makeChocolate(4, 1, 9) → 4
makeChocolate(4, 1, 10) → -1
makeChocolate(4, 1, 7) → 2[/COLOR]

Code:

public int makeChocolate(int small, int big, int goal) {
int remainder = goal%5;
int totalSum = small*1+big*5;
int miniSum = goal-(goal%5)+small;

if(totalSum<goal) {
    return -1;
}
else {
    if((big*5>goal) && (miniSum<goal)) {
       return -1;
    }
}
    
if(big<goal/5) {
    return (goal - big*5);
}
else {
    return remainder;
}
}

Stereo Wrote:

 Spoiler

To me, if the question implies an ordering to the process, it's better if the code mimics that process. Start by subtracting as many large chocolate bars as possible from the total, then consider whether the small ones are enough to fit the remainder. If you separate the two steps like that, it makes it easier to read what's happening than if you calculate a few numbers at the start, then compare them and spit out a result.

In this case, you want to subtract either
a) all the large chocolate
b) as many large chocolates as fit in the total
Depends which is smaller.
-- a is easy to calculate, it's just 'big'
-- b is total/5
So,

Code:

// remove as many big chocolates as possible, while leaving total positive
if (total/5 < big)
  total = total%5;
else
  total = total - 5*big;

Then compare the remaining total to the number of small chocolates you have available.

Code:

// return the number of small chocolates necessary to make up the total
if (small >= total)
  return total;
else
  return -1;

By doing it this way I've also reduced your 4 if conditions to 2.