Tip 1 : Rather than learning new things first try to practice and grasp already learned thing.
Tip 2 : Always try to understand the concept before grasping anything.
Tip 3 : Never copy your project.
Tip 1 : Always include only those things in which you are good.
Tip 2 : Don't try to add so many languages (programming)
Timings were good I mean mine were at 5 PM some people also had at 8Pm but I guess no one had more late than this.
As the test was online so we had to sit in calm environment.
yes there was one round of interview and it was quite hard.



Given 'N' : 5 (number of packets) and 'M' : 3 (number of students)

And chocolates in each packet is : {8, 11, 7, 15, 2}
All possible way to distribute 5 packets of chocolates among 3 students are -
( 8,15, 7 ) difference of maximum-minimum is ‘15 - 7’ = ‘8’
( 8, 15, 2 ) difference of maximum-minimum is ‘15 - 2’ = ‘13’
( 8, 15, 11 ) difference of maximum-minimum is ‘15 - 8’ = ‘7’
( 8, 7, 2 ) difference of maximum-minimum is ‘8 - 2’ = ‘6’
( 8, 7, 11 ) difference of maximum-minimum is ‘11 - 7’ = ‘4’
( 8, 2, 11 ) difference of maximum-minimum is ‘11 - 2’ = ‘9’
( 15, 7, 2 ) difference of maximum-minimum is ‘15 - 2’ = 13’
( 15, 7, 11 ) difference of maximum-minimum is ‘15 - 7’ = ‘8’
( 15, 2, 11 ) difference of maximum-minimum is ‘15 - 2’ = ‘13’
( 7, 2, 11 ) difference of maximum-minimum is ‘11 - 2’ = ‘9’
Hence there are 10 possible ways to distribute ‘5’ packets of chocolate among the ‘3’ students and difference of combination (8, 7, 11) is ‘maximum - minimum’ = ‘11 - 7’ = ‘4’ is minimum in all of the above.
This round was quite difficult they asked a lot of questions reagarding projects. Questions were also asked regarding web sockets but that was also part of my project so the only thing I want to say is for interview have a good knowledge of project.



• The left subtree of a node contains only nodes with data less than the node’s data.
• The right subtree of a node contains only nodes with data greater than the node’s data.
• Both the left and right subtrees must also be binary search trees.
For the given BST, if we want to delete the node with data 5 :

The modified BST will be:

Please notice that another valid answer is:

1. The node which we want to delete will always be present in the given tree.
2. If after deletion the tree becomes empty, print -1.
If the root is NULL, then return root (Base case)
If the key is less than the root’s value, then set root->left = deleteNode(root->left, key)
If the key is greater than the root’s value, then set root->right = deleteNode(root->right, key)
Else check
If the root is a leaf node then return null
else if it has only the left child, then return the left child
else if it has only the right child, then return the right child
else set the value of root as of its inorder successor and recur to delete the node with the value of the inorder successor
Return

Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
What is recursion?