According to me, there is a stack of n cards. And a value of q is given. I have to do q iterations and in every iteration, we will take out i-1 cards and see the ith card. If this ith card number is divisible by the ith prime number, then we will put it into another stack of cards.
And if not divisible, the card would be in the same pile/ stack.
And after every iteration, we are also placing back the removed i-1 cards.
am i right?
Hello @sirraghavgupta,
I would say you are partially correct.
N is the no. of cards in A(0)th stack.
Q is the no. of iterations you have to perform.
The third input is N numbers, representing the number of cards present in A(0)th stack.
Let’s understand this with the help of an example:
INPUT:
N=5
Q=2
A(0)th stack : 1 2 3 4 5
Processing:
iterate for i =1 to Q;
for i=1(frst iteration):
Pop top element of A(i-1)th (A(0)th) stack i.e. 5
Check, ((top_element i.e. 5)%(ith prime number i.e. 1st prime number, 2)) == 0 ? Push top_element to B(i) i.e. B(1)th stack : Push top_element to A(i)i.e.A(1)th stack;
After repeating the same for all elements of stack A(0):
A(0)=[] i.e. empty as all elements have been popped.
A(1)=[5,3,1] as they are not divisible by 2.
B(1)=[4,2] as they are divisible by 2.
for i=2=Q (second and last iteration):
Pop top element of A(i-1)th (A(1)th) stack i.e. 1
Check, ((top_element i.e. 1)%(ith prime number i.e. 2nd prime number, 3)) == 0 ? Push top_element to B(i) i.e. B(2)th stack : Push top_element to A(i)i.e.A(2)th stack;
After repeating the same for all elements of stack A(1):
A(1)=[] i.e. empty as all elements have been popped.
A(2)=[1,5] as they are not divisible by 3.
B(2)=[3] as they are divisible by 3.
Output:
print elements of B(1),B(2),A(2)
2
4
3
5
1
Note: top element of the stack is popped first.
Hope, this would help.
Give a like if you are satisfied.