Tip 1: Practice questions from all the topics as much as possible.
Tip 2: Be very attentive while working on projects in college or elsewhere, as they will be asked about in detail if mentioned on your resume.
Tip 3: Be consistent while practicing and try a variety of questions rather than doing many questions of the same kind.
Tip 1: You should know everything you have mentioned in your resume. Never brag in your resume. Be very precise. It doesn’t matter how much you have done; what matters is that you are very confident and clear about what you have accomplished.
Tip 2: Make sure to include your projects and the programming languages you code in on your resume.
The test had 28 mcqs and 2 coding questions.
The test was proctored. One needed to have webcam on.
A sample test link was provided before test to get familiar with the mettl platform.



A pair ('ARR[i]', 'ARR[j]') is said to be an inversion when:
1. 'ARR[i] > 'ARR[j]'
2. 'i' < 'j'
Where 'i' and 'j' denote the indices ranging from [0, 'N').
Brute Force Approach got submitted.
Approach:
I used and O(n^2) approach, where I used 2 for loops.
step1 : First loop was from index 0 to sizeof given array i.e. i
step2 : inside this was another loop from index of first loop + 1 to the end of array i.e. j.
step3 : If the value of array[i]>array[j] add 1 to the count of inversions.
Then, return count of inversions at the end of the for loops.
LiveCode : To write code, it was not a compiler.
The interviewer was quite supportive.



I solved this problem with the very first approach that can come to anyone's mind.
That is using recursion
Step 1 : Make a separate function to find middle element of a list using runner technique.
Step 2 : Find middle element of the list and make it the root of the BST and pass the left part of middle node of the list in the function to return the left node of the current node of BST and right part to get the right child subtree of the current node of BST.
Step 3 : Return root of BST.



1. Push(num): Push the given number in the stack if the stack is not full.
2. Pop: Remove and print the top element from the stack if present, else print -1.
3. Top: Print the top element of the stack if present, else print -1.
4. isEmpty: Print 1 if the stack is empty, else print 0.
5. isFull: Print 1 if the stack is full, else print 0.
We perform the following operations on an empty stack which has capacity 2:
When operation 1 1 is performed, we insert 1 in the stack.
When operation 1 2 is performed, we insert 2 in the stack.
When operation 2 is performed, we remove the top element from the stack and print 2.
When operation 3 is performed, we print the top element of the stack, i.e., 3.
When operation 4 is performed, we print 0 because the stack is not empty.
When operation 5 is performed, we print 0 because the stack is size 1, which is not equal to its capacity.
Stack implementation.

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