How will new stacks be created if we have multiple queries?
Explain Approach
Hello @ankush.bhardwaj0,
Lets understand the question, it will help you to figure out the approach:
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.
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