Problem 9. Who Has the Most Change?

Why not just have the coin values in ascending order? took a while to realise my stupidity fgs.

Read the order of the coins/all of the problem carefully, it's easy to miss a subtle switch

Nice problem

Nice problem!

Be careful about the order of the coins!

using martrix multiplitation make this problem easier.

You need to consider the order given in the problem description

test suit #5 and 6 are wrong

cool

Nice problem!
Read the description carefully!
The Oder of Coins is matter

This problem emphasizes our comprehension on the problem rather than our ability to code.

I think it checks both

test5 and 6 are false answer

yes,me too

then u should read the context carefully. be aware of the order of the change

This test fail =
a = [ 0 1 2 1; 0 2 1 1];
c = 1;
assert(isequal(most_change(a),c))

I don't undersdand why.
The first person = 0 * 0.25 + 1 * 0.10 + 2 * 0.05 + 1 * 0.01 = 0.21
The second person = 0 *0.25 + 2 * 0.10 + 1 *0.05 + 1 * 0.01 = 0.26

So the first person has less money than the second, should return 2.

I'm missing something here?

d = [0.25 0.05 0.1 0.01];
[c idx]= max(sum((a.*d)'));
b = idx;

This works but how to optimize it to reduce the size.

function b = most_change(a)
a(:,1)=a(:,1)*0.25;
a(:,2)=a(:,2)*0.1;
a(:,3)=a(:,3)*0.05;
a(:,4)=a(:,4)*0.01;
c=a(:,1);
d=[]
for i=1:length(c)
d(i)=sum(a(i,:))
end
[M,idx]=max(d)
b=idx
end

why does this not work?

a(:,2) should be multiplied with 0.05
a(:,3) should be multiplied with 0.1

since they are not ordered in the question
thank you

Look at the description carefully, there is a trick.
The correct value matrix is [0.25 0.05 0.1 0.01].

exactly !

Thanks

[~,b]= max(sum((a.*[.25,.05,.1,.01]),2))

a = [0 1 0 0; 0 0 1 0];
c = 2;

How is the second one the one with the highest money? The first one has $ 0.10 whereas, the second has 0.05.

a = [ 0 1 2 1; 0 2 1 1];
c = 1;

Here too! How is the first one higher? When you evaluate, you get $ 0.21 for the first one and $ 0.26 for the second.

Sneaky

It's a trap! read carefully.

tricky order

For this problem the two tests of the test suit have issue

Test5:
a = [ 0 1 2 1; 0 2 1 1];
c = 1;

Test6:
a = [0 1 0 0; 0 0 1 0];
c = 2;

The solution should be other way around.

Why is Code 5 and Code 6 so gross?

I don't see how the test problems 5 and 6 are correct, they are displaying the answer the lower monies.

it is so funny that the order is not exactly according to the "right order".
you need more seriously read the question.

Take care of the value of dime = $0.10, nickel = $0.05. The order in the matrix is just replaced. silly trick

the expected results of the test case 5 and test case 6 are not correct.

Clumsy version that seems OK in Octave, but then so did the earlier more elegant one

Jerks. Every fast food place I've worked (and there's been a bunch) orders the register sequentially.

Is this about coding Matlab or trick questions? If the rest of these are trick questions I'll go somewhere else.

I thought Cody was about testing your Matlab skill. This task with its deceitful wording is far from being that as I imagine most of the people got the code right but failed at Test 5 and 6 due to the purposeless and illogical coin order swap in the first sentence of the problem.

Thx, it took me a shit ton of time figuring that out. Every time I red the question I did not notice that bit so thx.

Tricky

Test 5 and 6 are wrong in their answers please check

Tips:in that order! in that order!!!

fell for it!

just look around for the solutions

mandovà

Test 5 and 6 combinations are not correct. Test 6 similar to the example given at the starting, but both answers are contradicting.

c'est nul

It's important to read carefully the description, maybe it should be ordered so to not create confusion.

why my procedure can not pass the test5 and test6

Yes,me too.

It's a trap!
read carefully.

very easy

gj

This is wrong or not?

Very good. It is very interesting to solve it. The most problem is to optimize the program and guarantee the small size.

Very sneaky!

Could anybody help guide me in the right direction?
The current code fails on the "c" outputs

"b" or "c" are just arguments of isequal() function. It does not matter.
As the tag says, "notice the rignt sequence of money!"

Test 5 and 6 change the variable and appear to be incorrect when denoting which person has more money. Ex. b = 2 b/c .26 is more than .21, but the answer shows c = 1

%my code runs perfectly in matlab
function b = most_change(a)
b = 1;
clc
quarter = 0.25;
dime =0.10;
nickel =0.05;
penny =0.01;
y = [quarter,dime,nickel,penny]';
out = a*y;
[u v] = max(out);
b = v;
end

Test suite #2 and 3 incorrect

They are correct.
This is a trick question.

As far as I can see it fails in 5 and 6 because the assertion answers provided are wrong. Furthermore in test 1-4 the assertion answer is called b then in 5-7 c

This is a trick question. The test 6 says that
% Now go back and read the problem description carefully.

Very deceptive question. see through the test suite before submitting

Point 5 and 6 works on my station with b=1 and 2. I don't understand what is the "c=1/2" in the solution ?

I'm happy to get to 1-line solution.

on Test 5 and 6, to be honest I cheated,because they dont check correctly!Shame on you MATLAB

Be careful the order is quarters, nickels, dimes, and pennies. Not quarters, dimes, nickels, and pennies

Start with the following...
a*[25 5 10 1]'
and you know how much each person has.

I dont know why my solution doesnt satisfy all the test inputs. If someone could explain what I did wrong that would be great.

Nickel comes before dime. Also, there is a five person test, so you need k=5.

Not sure the test suite is correct on this problem. If I have 1 dime, two nickels and a penny then I will have (21 cents). If I have two dimes, a nickel and a penny, I will have 26 cents. So the second is larger, but the test suite is testing that the row should be 1. Both 5 and 6 seem off.

Hello I am a South African newbie, and this is how I solved it

function b = most_change(a)

a(:,1) = 0.25 .* a(:,1);
a(:,2) = 0.05 .* a(:,2);
a(:,3) = 0.1 .* a(:,3);
a(:,4) = 0.01 .* a(:,4);

a = a';

vecsum = sum(a);

vecsum = vecsum';

[M,I] = max(vecsum);

b = I;

end

Almost had it!! The ordering has nickel and dime switched. So just change v to
[.25 .05 .1 .01] and you're done.

it has failed for a = [ 0 1 2 1; 0 2 1 1] , says 1st person has more money. How is 0.21>0.26 ??

Read the problem carefully in terms of the order of the coins.

Hm, I did not read the descrption that carefully the first times ;)

test 5 and test 6 are wrong.
the true answer of test 5 is 2, and for test 6 it is 1.

You have inverted dime and nickel in your problem wich produce an inconvenient mystake.

Good, but the little trick felt a little underhand

Problem is good, but tet cases are contradict to conditions.

That was a dirty trick, lol. It was a forehead slapper once I saw it.

this runs well on matlab and octave. Am I missing something?

"n_rows=size(a)(1,1)" should read "n_rows=size(a,1)"

Thank you! I'm getting confused between this and octave syntax:)

Test 5;
a = [ 0 1 2 1; 0 2 1 1];
c = 1;

a(1, :) = 0*.25+1*.10+ 2*0.05 + 1*0.01 = $ 0.21
a(2, :) = 0*.25+2*.10+ 1*0.05 + 1*0.01 = $ 0.26

so, output should be 2...

Similarly, for Test 6 is wrong

a(1, :) = $ .10;
a(2, :) = $ .05;

so output should be 1

not entirely sure why my code fails for test#5 & 6..

test#5 & 6 are wrong

poorly written problem

Definitely agree with Jacob too!

Thanks for the feedback. We will discuss and see if we can ban the use of regexp

Thanks for your attention, Aditya Jain :-)

function ans = most_change(a)
m=[0.25 0.10 0.05 0.01];
[r,c]=size(a);
for i=1:r
c(i)=sum(a(i,:).*m);
end
for j=1:r
if(max(c)==c(j))
ans=j
end
end

all are correct but fails 5 ,6 test suit pls help me

Read the comments for the problem.

To the one who made this problem:you are not really a very good person...O(∩_∩)O

The author is not a good person.....

Nice work!!

Can you explain how this works please?

This is only a (shameless) trick to save one cody point... hint: double('á-Z ')=[225 45 90 9]=9*[25 5 10 1]

Brilliant idea! How did the conversion to character even occur to you?

in Test,case 5 and 6 is wrong,a=[ 0 1 2 1; 0 2 1 1],c should be 2,that means person 2 has more money

Watch out, they've inverted the dimes and nickels in the input vector!

The test #5 is incorrect...
a = [ 0 1 2 1; 0 2 1 1];
c = 1;
but the second one has more money...

Also test #6 is incorrect...a = [0 1 0 0; 0 0 1 0]; c = 2; but the first one has 10 cents and the second one has only 5...

Thank you, but wouldn't that mean that the apostophes are unbalanced (which for some reason they are not) -> 3 apostrophes, 1 transpose?

The 1st and 4th apostrophes denote the beginning and end of the character string, and the 2nd and 3rd represent a single apostrophe inside the string. A pair of apostrophes together is what you use to get one apostrophe inside a string. It's just one of those awkward computer things.

Thanks a lot, Tim.

Doh! I should have known that this is a matrix multiply!

Annoying obfuscation regarding the input variable's coin 'column ordering'. Shame on you, Cody.

Any ans for Neural Nomad's question. Pls give some reasoning.

str2num evaluates to that vector, so just putting in the vector works also. But Cody just thinks that this code is "shorter": evaluating a function with one argument is less nodes than concatenating 4 values.

Although I appreciate the effort, "size" calculation is B.S. It should calculate time it takes to run the function.

the last suite test is flawed

test 5 is wrong.
Not yet corrected.

The final test case solution is incorrect.

The test # 5 is still wrong.

Test 5 is wrong, (c=2)

Is test case 5 correct? 1 -> 0.21, 2 -> 0.26!!

No 5. Test suite is wrong. For the matrix
[0 1 2 1; 0 2 1 1]; the correct answer is 2 and not 1 as indicated in the test suite.

Why does the second row have more change than the first in

a = [ 0 1 2 1; 0 2 1 1];

Does this setup not mean that row 1 has
0*0.25 + 1*0.05 + 2*0.1 + 1*0.01=0.26
whereas row 2 has 0*0.25 + 2*0.05 + 1*0.1 + 1*0.01 = 0.21 < 0.26?

What am I missing?

You did not misread. The order still appears with nickels before dimes at the point where the order is specified (not in the list where values are given).

Test case #5 should have b = 1. This is rather annoying. In addition to the "Remove consonants" problem actually looking for removal of vowels, Cody needs some serious work before it's ready for prime time.

I fixed the last test case and pushed a re-score. Very sorry for the confusion.

:))

why does this has a lower score than when using [0.25 ; 0.10 ; 0.05 ; 0.01] ???

The test suite wasn't sufficiently rich. I've added some tests, so now this fails.

wrong solution, says which person has the most coins, not the most money

Ditto, it appears the test case(s) don't catch this one.

Yes, this solution returns the person who has the most coins, not the most change.

I've added to the test suite to help close this loophole.