This code is giving run error for two test cases

#1


constraints for q is not given

#2

Hello @ankitdelhi20,

Your code will run only for q=1.

  1. You have to write a generalized code irrespective of q’s value.
    Solution:
    Create two arrays of the stack of size q+1:
  • One is for A
  • another for B
  1. You are computing prime numbers up to 10 which may cause any error.
    Make it 10^9 to cover the entire range of integer as it is not specified in the question.

Let’s understand the question first:

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, repeating the number of cards present in A(0)th stack.

Approach:
You have to run a loop Q times to check for every ith prime number.

for(int i=1;i<=q;i++)
{
/*

For each value of i, process A(i-1)th stack.
While processing, check if the elements of A(i-1)th stack are divisible by ith prime number or not.
if it is divisible add that element/number to B(i)th stack(B is another set of stacks).
If not then push that element to A(i)th stack.
*/
}
Print all the stacks of B i.e. from B(0) to B(Q) and then A(Q)th stack at last.

Note: elements in stack are pushed and popped in a LIFO manner.

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.
If you still have doubts, feel free to ask.

#3

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