Correct Change I Comments

Comments

  • For input set 5 3 19, we can get 19 using 3 pennies and 4 nickels. So it should be true there.

  • Sorry, I mistook nickels and pennies...

  • 5, 4, 23 will result false !!!!!! it ca't be true 5*5 +4 = 24 ?????????????????

  • @Amboss, the challenge is to see if you can select the exact change from the coins you are given. So if you have 5 nickels, you can select 4 of them, add 4 pennies for a sum of 24 cents.

  • I've made the code, but when I run table below shows that all cases are correct EXCEPT that "Your Output" is incorrect, it's red.
    I don't know why, I checked some made up cases manually on Eclipse.

  • Кирилл, When given 7 nickels and 3 pennies, can you get 35 cents? Yes, you just select the nickels and ignore the pennies. Check your code to see what it returns.

  • Admin (?), I've make some changes, even few, but still, there is something wrong. Your example works, thanks for it!)
    I'm just curious where I get lost.

  • Кирилл. there's another example you're getting wrong. I think you should approach this problem a little differently. You can solve it with one if-else statement.

  • You can solve it with one if-else statement.
    You're right. And only now it's shows "Correct". I have to say that with no mathematical help I won't do that. So, on my opinion this problem more related to mathematics rather then to programming.

  • OMG :( this took me forever also my code is extensive.... looks like complying with DRY is going to take me looong time to master.

  • I solved !!
    if (sum / 5 <= nickels && sum % 5 <= pennies){
    return true;
    }
    else {
    return false;
    }

  • At the end of my program I write << return boolean; >> but program say error:

    variable bool might not have been initialized
    return bool;

  • I found my solution;
    when I defined boolean bool I must give amount to it LIKE boolean bool = false;

  • I solved it but some complicated:
    boolean bool=false;
    for (int i =0; i<=pennies; i=i+1){
    if((sum - i)%5 == 0)
    { bool= true; break;}

    }           return bool;
    
  • boolean statement worked for me
    boolean bob = sum % 5 <=pennies;
    return bob;

  • boolean bob = sum%5<=pennies;
    return bob;
    
  • if(sum%5 > pennies) {return false;}
    return nickels*5 + pennies >= sum;

  • I archived many comments here with solutions. You can link to your solution (in your submissions) to share it, but you shouldn't paste the whole solution in here.

  • @mistermase nice. See also the featured answer

  • i'm confused i'm not familiar with nickel and pennies and cents -.- do they have relationships with each other ? like how many cents are there in pennies or in nickel ? :)

  • i'm really sorry :( we have different currency here in our country .... that's why i'm confused -.-

  • American Currency^ LOL

  • For those of you in different countries: Pennies are 1 cent, Nickles are 5 cents (each).

  • Added a note to main content on their value.

  • I think i have a good solution? is it a smart solution?

    my code

  • OK, but see the featured answers for a more concise answer.

  • I wrote my code in python. it's working if i run this in some other compiler but here it's not.
    my code

  • Boilerplate code for this challenge wasn't provided for Python, so you need to take in the input yourself from STDIN.

  •     int nCount = sum / 5;
        if (nCount <= nickels) {
            return sum % 5 <= pennies;
        } else {
            return false;
        }       
    
  • int a=sum/5;
    int b=sum%5;
    if(nickels>=a && pennies>=b){
    return true;
    }
    return false;

  • if ((sum%nickels) > pennies) return false;
        return true;
    

    how the heck did this pass all the tests..

  • Thanks, that's actually an edge case that wasn't covered before. The last test case will now check that condition so that code no longer passes.

  • I did this way
    if(nickels>=sum/5 && pennies>=sum%5)
    {
    return true;
    }
    else
    {
    return false;
    }

  • Here's some working boilerplate for Ruby:

    n = gets.chomp.to_i
    n.times do
    input = gets.chomp.split(" ").map(&:to_i)
    nickels = input[0]
    pennies = input[1]
    desired_sum = input[2]
    print correct_change?(nickels, pennies, desired_sum), "\n"
    end

  • here goes my take!

    if (sum % nickels <= pennies && ((nickels *5)+ pennies) >= sum){
    return true;
    }
    else{
    return false;

Contact Us
Sign in or email us at [email protected]