Its olsom doen't work, could you look at it?

You have coded in c language

Compare The Triplets problem is very simple and similar to comparing marks of three subject of two students in a way like if subject1, student1 got high marks than student2 than student1 got 1 point and the same repeat for subject2, and subject3 at the end print the both student points. Maximum points can a student earn is 3 or minimum can be 0 or there may be the equal point if in subject1 both students got the same point. Check the explanation section for better understanding.

Explanation:- Let's take an example and compare with our logic, suppose two students are Bob and Alice and Bob got 56, 75, 26 marks and Alice got 84, 45, 26 Now Compare The Triplets

Bob Marks===> 56, 75, 26 Alice Marks==> 84, 45, 26

Bob Marks Alice Marks Bob Point Alice Point

56 < 84 0 1 75 > 45 1 0 26 = 26 0 0

So we can see that Bob and Alice got 1, 1 point in tale That is an answer print on the screen. Read Full Article Compare The Triplets

var is a dynamic type variable that the compiler infers the type based on the assignment. Can't be used in a method signature.

thanks bro

www.ask2exp.com sie does not has ssl certificate

I couldn't find any solution at this website. Can you tell us how?

I have created what might be the worst possible functional solution:

int compare(int a, int b)
{
    if (a > b)
    {
        return 1;
    }
    else if (b > a)
    {
        return -1;
    }
    else
    {
        return 0;
    }
}

vector < int > solve(int a0, int a1, int a2, int b0, int b1, int b2){
    int a = 0; // Difference of A and B's scores
    int b = 0; // Number of points awarded in total
    int c = 0; // Current return value
    int d = 0; // This is A's score
    int e = 0; // This is B's score
    c = compare(a0,b0);
    a += c;
    b += abs(c);
    c = compare(a1,b1);
    a += c;
    b += abs(c);
    c = compare(a2,b2);
    a += c;
    b += abs(c);
    d = a + (b-a)/2;
    e = b - d;
    vector < int > retvals = {d, e};
    return retvals;
}

Good intentions lead me to bad code

great great great work... that is like a programmer...

thanks bro its right approach..

but my logic

static int[] solve(int a0, int a1, int a2, int b0, int b1, int b2){ int alice = 0; int bob = 0;
if(a0!=b0){ int c0 = ((a0>b0)?1:0); switch(c0){ case 1: alice++; break; case 0: bob++; break; default : break; } } if(a1!=b1){ int c1 = ((a1>b1)?1:0); switch(c1){ case 1: alice++; break; case 0: bob++; break; default : break; } } if(a2!=b2){
int c2 = ((a2>b2)?1:0); switch(c2){ case 1: alice++; break; case 0: bob++; break; default : break; } } int [] scor = { alice, bob }; return scor; }

such a bad code

I would recommend to use one single segment of code for same functionality. Don't repeat your self.

Realy clean code and more easy to understand. Awesome job mate.

u can use while loop to simplify the code

This is really bad code. In programming you should always have extending in mind when writing code. Now imagine you have to compare one hundred grades, with your approach you would have to use 200 variables

not really readable

Far more elegant and simple python3 solution.

def solve(a0, a1, a2, b0, b1, b2):
    arr = [a0, a1, a2, b0, b1, b2]
    return compare(arr, 0, len(arr) - 1, [0, 0])
    

def compare(a, start, stop, result):
    if start >= stop:
        return str(result[0]) + str(result[1])
    elif a[start] > a[stop]:
        result[0] += 1
    elif a[start] < a[stop]:
        result[1] += 1
    return compare(a, start + 1, stop - 1, result)

heyy not executing man

I had a similar solution! Just to be a little nitpicky you can make the second "if" statement within the for loop as an "else if" so that the condition doesn't get evaluated if the first one is true already.

Similar solution, just make sure you make your programs dynamic so that you have minimilistic hardcoding. I hope you will understand comapring below and your's

why faster and scaleable? Linked list faster than array list, use of else if.

scalable because I haven't specified the hardcoded length for inputs

static List<Integer> compareTriplets(List<Integer> a, List<Integer> b) {
    int aWin = 0;
    int bWin = 0;

    List<Integer> win = new LinkedList<Integer>();
    for (int i = 0; i < a.size(); i++) {
        if (a.get(i) > b.get(i))
            ++aWin;
        else if (a.get(i) < b.get(i))
            ++bWin;

    }
    win.add(aWin);
    win.add(bWin);
    return win;
}

Boa solução mas não contempla a restrição de >= 1 e <= 100.

Why is this downvoted ?

Thank you so much for explaining this concept. Your expalanation is very clear and easy to understand. Thank you for your time.

Thanks for making it understandable.

Thank you for explaining the concept of Ternary operators.

Im confused, for the other answers how I'm reading this would work is essentially;Say these are both going to equal true, if (a0 > b0) you get aliceScore = 1. And then when you go to the next if (a1 > b1) you get aliceScore = 1 again. Instead of aliceScore += 1; Can you help me understand that part. Basically I'm not understanding where is the (if true +1), I'm only seeing (if true set it to 1).

sweet explanation of the syntax. Thanks.

How  this  particular  code  set  works  in  c?

Ternary operator is same as if - else if we use ternary operator we can write the if(condition) do this; else do this; in same line using condition ? true do this : false do this

We use them. While the ternary operator is nice and concise in this particular solution, it will not scale as soon as you add more data. Just imagine the code you'd have to write if it turned out that its 10 or 100 scores that you need to compare - instead of 3.

Ternary operator is same as if - else if we use ternary operator we can write the if(condition) do this; else do this; in same line using condition ? true do this : false do this

These are ternary operators, so (a0>b0)?1:0 is essentially saying "evaluate a0>b0, if it's true evaluate to 1, otherwise 0". Basically this is the form:

(boolean expression)? answer if true : answer if false

Facepalm myself when i saw your answer...

def solve(a0, a1, a2, b0, b1, b2):
    #
    # Write your code here.
    #
    a = [a0,a1,a2]
    b = [b0,b1,b2]
    
    alice = [x for x in list(map(lambda x,y: x>y, a,b)) if x==True]
    bob = [x for x in list(map(lambda x,y: x>y, b,a)) if x==True]
    
    return len(alice),len(bob)

Is your function returning a string? If so, this is the issue. The join method expects an array, so you want to be returning something in the format [aScore, bScore].

Then, result.join will convert the array into the expected output format.

This at least worked for me, hope my explanation is of some use to you

You don't have to, if it is only one line statement~ =D if there are multiple line, then you will have to put {}. For example:

if (a == 0)
    c++;

else if (a == 1)
    c--;

else if (a ==2)
{
    b++;
    b+c;
}