CHEFTET July Challenge help

Can Somebody help me about what’s wrong with my solution for JULY challenge problem chef and tetris and my approach was to check out 4 possible values so that we can make all array elements equal but I am getting wrong answer for 7th and 12th task

my solution :

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can you tell me what is wrong in my solution for chef and tetris.


my approach is somewhat similar to you. i also try to make all element equal to a[0] or a[0]+b[0] or a[0]+b[1] or a[0]+b[0]+b[1]. i try to go through your code what its hard to get what you are trying to do. one funny thing is that your solution is not passing the 7th and 12th task and my solution is failing on 1st task.

@sau1999 I tried running your code on my PC, for this particular input:
1 2 3
1 2 3
your code gives output:
but the correct output should be -1, in your code, I think you’re using the array ‘vis’ to check if elements in ‘b’ have been used or not i.e vis*==0 indicates that the 'i’th element has not been used and vis*==-1 indicates that the 'i’th element has been used. I have added this line:
cout << vis[1] << " " << vis[n-1] << " " << c* << endl;
just before the inner for loop is closed,I’ve noticed that vis[n-1] is ‘0’ for all the four iterations, and yet your code gives an output ‘3’, basically you’re not checking if the last element in ‘b’ has been used or not i.e if vis[n-1]==0.

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@hemanth_1 please tell me why my solution is wrong.


This happened with me as well.
It is happening because of n=1.

Just a separate if statement for n=1 needs to be added.

my wrong solution

my correct solution

your corrected solution

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@prakhariitd thanks

Please tell me what’s wrong with this :

Thanks in advance.

Can anyone please tell me whats wrong in my code

it gives wrong answer on Task #9


Hey… in this problem i got stuck in the 10th test case… can any one tell whats wrong with my approach??
my algo is:
if all elements of (a[n]+b[n]) is not divisible by n–> then print -1
else check for all the possible sums for one a* { b[i-1]+a* , b*+a* , b[i+1]+a* , b[i-1]+b*+a* , b[i+1]+b*+a* ,b[i-1]+b[i+1]+a* and b[i-1]+b*+b[i+1]+a*} from these if a*+any of these are equal to the desired element that is (a[n]+b[n])/n then making the/those b* value/values zero i proceed to check next a* element.
here is the source code…

@hemanth_1 : thanks for giving answer but i think for your given test case answer should be 3 because
b = {1,2,3}
a = {1,2,3}
b[1] will fall on a[0] (2 + 1 = 3)
b[0] will fall on a[1] (1 + 2 = 3)
and as a[2] is already 3 no element will on a[2] thus vis[n-1] = 0`

@sau1999 Where shall b[2] fall?

@sau1999 We have to use all the elements of array b.

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All the elements in ‘B’ must be used

thanks for clearing the problem statement I asked this question on the contest but no one replied