Please correct my code

help
, ,
1

I wrote my code and it works perfectly well when running it, but it shows a problem when i submitted the Question. Please help me re write my code

t=int(input())
for i in range(t):
    n,x,y=map(int,input().split())
    a= y//3 
    l= (x-a)//2
    f= l+a
    if l>=0:
        if f>=n:
            print("Yes")
        else:
            print("No")
    else:
        print("No")

Problem Link: Decorating Christmas Tree Practice Coding Problem - CodeChef

2

I wonder if this is actually an 1100 problemā€¦ ANYWAY!!

Your issue is that you are not optimizing your sources (large ornaments: x)

Since X is your source that is most likely to dispose, try to use your Y first.

Each 3 Y need 1 X, so:

one = min(X, Y//3)

After emptied your option one possibilities, emptie your second (considering the one you already used)

two = (X-one)//2

Finally, this:

# cook your dish here

#from collections import deque,Counter
#from math import ceil, floor

def innum():
    return int(input())
def inmany():
    return map(int,input().split())

for T in range(innum()):
    
    N,X,Y = inmany()
    
    one = min(X,(Y//3))
    two = (X-one)//2
    
    if one+two >= N:
        print("YES")
    else:
        print("NO")
1 Like
3

But arnt both the codes same? I mean i do agree that your code is better structured, but i dont understand where i went wrong with my approach in this question(My logicā€¦)

4

Nope, we are not using the same logic :sweat_smile:

I am using an min() function, your question should be asking me why I did that.

When I do this:

one = min(X,Y//3)

Means: ā€œI will reduce Y, but I if donā€™t have enough X to make an option one ā€˜kind one ornamentā€™ then option one is 0ā€.

Your code does not do that.
You code directly assumes there will be enough X to make Y//3.

1 Like
5

Ahh i understand now, thanks a lot!!!

6

Iā€™m glad :hugs:

If any wonders are left, let me know.

7

:+1: