Tip 1 :Prepare your resume with good projects and keep it short , add project links, and keep it simple
Tip 2 : Practice daily will increase your confidence and skill as well
Tip 3 : Give daily contest on leetcode, CodeStudio will Introduce you new questions
Tip 1: Keep it genuine and simple
Tip 2: Atleast 2 projects .
It was on 12 pm around. I have to attend the test with camera and microphone on in online mode.



I Calculate the MAX value in the array.
I Create an array of size MAX and store the occurrences of each element in it.
Since we want to maximize our answer, we will start iterating from the MAX value to 0.
If the occurrence of the ith element is greater than 0, then add it to our answer decrease the occurrences of the i-1th element by 1, and also decrease the occurrence of ith by 1 since we have added it to our answer.
We don’t have to decrease the occurrence of the i+1th element because we are already starting from the end so i+1th is already processed.
There might be multiple occurrences of the ith element that’s why do not decrease i yet, to stay on the same element.



For each node there can be four ways that the max path goes through the node:
Node only
Max path through Left Child + Node
Max path through Right Child + Node
Max path through Left Child + Node + Max path through Right Child
The idea is to keep track of four paths and pick up the max one in the end. An important thing to note is, that the root of every subtree needs to return the maximum path sum such that at most one child of the root is involved. This is needed for the parent function call. In the below code, this sum is stored in ‘max_single’ and returned by the recursive function.



The given linked list is 1 -> 2 -> 3 -> 4-> NULL. Then the reverse linked list is 4 -> 3 -> 2 -> 1 -> NULL and the head of the reversed linked list will be 4.
Can you solve this problem in O(N) time and O(1) space complexity?
Initialize three pointers prev as NULL, curr as head, and next as NULL.
Iterate through the linked list. In a loop, do the following:
Before changing the next of curr, store the next node
next = curr -> next
Now update the next pointer of curr to the prev
curr -> next = prev
Update prev as curr and curr as next
prev = curr
curr = next
I have done this by reversing its pointers.



For the given binary tree

The level order traversal will be {1,2,3,4,5,6,7}.
I just use bfs to solve the problem.
Create an empty queue q and push root in q.
Run While loop until q is not empty.
Initialize temp_node = q.front() and print temp_node->data.
Push temp_node’s children i.e. temp_node -> left then temp_node -> right to q
Pop front node from q.
It was an Hr round of 10 minutes
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Rate yourself on scale of 10 about your confidence ?
Tip 1:Always keep calm and confident

Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
How do you remove whitespace from the start of a string?