Tip 1 : Consistency
Tip 2 : Never Give Up
Tip 3 : Learn from the mistakes
Tip 1 : Simple and short providing the required information
Tip 2 : Single page resume with showing all the multiple projects which have done.
Online MCQ Test on Calyxpod platform:- Questions from Aptitude, OS, DBMS, DSA
Set of 50 simple MCQs on C++, aptitude, and logical. Only those who cleared the objective round were allowed to sit for the subjective round.
In this round, we had to write the whole program along with appropriate comments. The code should contain explanations wherever required. Many students who wrote the correct code and passed all test cases were not shortlisted because they forgot to write the comments. You could even include some edge cases in the comments and how they are being handled.
The majority of questions were asked on trees and linked lists. (and a little bit of DP). There was a total of 3 technical rounds followed by an HR Round. In each technical round, we were asked 2 questions. Sometimes if the candidate is unable to come up with the solution to any of the problems then another question is added (depends on the interviewer’s mood).



Input: Let the binary be as shown in the figure:
Output: Linked List: 15 -> 40 -> 62 -> 10 -> 20 -> NULL
Explanation: As shown in the figure, the right child of every node points to the next node, while the left node points to null.
Also, the nodes are in the same order as the pre-order traversal of the binary tree.
A simple solution is to use Level Order Traversal using Queue. In level order traversal, keep track of previous node. Make current node as right child of previous and left of previous node as NULL. This solution requires queue, but question asks to solve without additional data structure.
Recursively look for the node with no grandchildren and both left and right child in the left sub-tree. Then store node->right in temp and make node->right=node->left. Insert temp in first node NULL on right of node by node=node->right. Repeat until it is converted to linked list.



If the given sequence ‘ARR’ has ‘N’ elements then the sorted wave array looks like -
‘ARR[0] >= ARR[1]’ and ‘ARR[1] <= ARR[2]’
‘ARR[2] >= ARR[3]’ and ‘ARR[3] <= ARR[4]’
‘ARR[4] >= ARR[5]’ and ‘ARR[5] <= ARR[6]’ And so on.
1. ‘ARR[0]’ must be greater than or equal to ‘ARR[1]’.
2. There can be multiple arrays that look like a wave array but you have to return only one.
3. We have an internal function that will check your solution and return 'True' in case your array is one of the solutions otherwise return 'False'.
The given array ‘ ARR = { 4, 3, 5, 2, 3, 1, 2 } ’
The below figure is a visual representation of the given ‘ARR’ and you can see we can express ‘ARR’ in a waveform array because
4>3 and 3<5
5>2 and 2<3
3>1 and 1<2
And it follows the condition of wave array.

Try to solve this problem in linear time complexity.



All the elements in the array are distinct.
Input: arr = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] and it was rotated 3 times.
If we take closer look at examples, we can notice that the number of rotations is equal to index of minimum element. A simple linear solution is to find minimum element and returns its index.
Time Complexity : O(n)
Auxiliary Space : O(1)
Here also we find the index of minimum element, but using Binary Search. The idea is based on the below facts :
The minimum element is the only element whose previous is greater than it. If there is no previous element, then there is no rotation (first element is minimum). We check this condition for middle element by comparing it with (mid-1)’th and (mid+1)’th elements.
If the minimum element is not at the middle (neither mid nor mid + 1), then minimum element lies in either left half or right half.
If middle element is smaller than last element, then the minimum element lies in left half
Else minimum element lies in right half.
Time Complexity : O(Log n)
Auxiliary Space : O(Log n)
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How do you remove whitespace from the start of a string?