Tip 1 : Be Strong with DSA
Tip 2 : Be Strong with OOPS and Core CS Fundamentals
Tip 1 : One Page Resume
Tip 2 : Highlight your coding profiles
I only remember 1 coding question from this round. Basic level questions were asked.



1. If ‘k’ is not present in 'arr', then print -1.
2. There are no duplicate elements present in 'arr'.
3. 'arr' can be rotated only in the right direction.
Input: 'arr' = [12, 15, 18, 2, 4] , 'k' = 2
Output: 3
Explanation:
If 'arr' = [12, 15, 18, 2, 4] and 'k' = 2, then the position at which 'k' is present in the array is 3 (0-indexed).
Efficient Approach: To solve the problem, the idea is to use Binary Search based on the approach discussed in the article search an element in a sorted array. Follow the steps below to solve the problem:
Compare X with the middle element.
If X matches with the middle element (arr[mid]), return the index mid.
If X is found to be greater than the arr[mid], then X can only lie in the subarray [mid + 1, end]. So search for X in the subarray {arr[mid + 1], .., arr[end]} .
Otherwise, search in the subarray {arr[start], …., arr[mid]}
DSA


If the given matrix is:
[ [1, 2, 5],
[3, 4, 9],
[6, 7, 10]]
We have to find the position of 4. We will return {1,1} since A[1][1] = 4.
Algorithm:
Let the given element be x, create two variable i = 0, j = n-1 as index of row and column
Run a loop until i = n
Check if the current element is greater than x then decrease the count of j. Exclude the current column.
Check if the current element is less than x then increase the count of i. Exclude the current row.
If the element is equal, then print the position and end.
Project Discussion and General Behavioural Round
Introduce yourself.
What are your strengths and weaknesses?
Discuss your project with me.
Tip 1 : Make sure you know about your projects end to end

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