SRE 2

We can solve this questions with 2 approaches one was brute force where we just defined variable as product and we created two for loop and one for loop was to choose every value in an array and second was to choose the next value and we can check the product with every value if the product is divisible by 2 we can return the count of that no 
Second i gave the optimised code approach where we can count the total no of odd values in an array and we can directly subtract the total n - count of odd values in an array which will directly give the no of event product inside an array .

 

I solved this question with bfs approach created a queue which was initially point to the root values for the binary tree and after that start moving towards left and right one did the sum for left and right values in the tree and store it in vector and then move to the left and store the values and we can continue the process like that .

 

We create adjacency lists we can push inside the values into that adjacency lists and post that we can create a bool condition to check that if the graph is cyclic or not

Tip 1:We need to create the dockerfile and whatever configuration he want to add we can add we can add by creating stages 
Tip 2: command should be on your tips so that you can easily write the codes .
Tip 3: your explaining process should also be good so that you can explain things clearly .

Tip 1: we need to know how to create ingress.yaml
Tip 2: all the code specifications need to be very much cleared from your end 
Tip 3: port knowledge is very important