Tip 1 : Do not rush for hard topics be prepared with basic
Tip 2 : They mainly look for your basic knowledge so make strong roots in that.
Tip 3 : Be prepared with famous algorithms and their code also complexity.
Tip 1 : Clean and short
Tip 2 : Have some projects on your resume.
It was a simple round just an introduction and asked some data structure-related questions, and DBMS also
it was morning time and I get a slot from 11 am to 11.30
1. What is a list
2 . What is a pointer
3. Does java use a pointer
4. What are SQL and NoSQL
5. What you know about reliance jio
6. What is the OOPS concept
7. DBMS concept ( primary key, 2 vs 3 tier architecture, ACID properties )
8. The logic for basic loops Is there any way to print 1 to 100 without using any loops Also the interviewer started to defend their side and refuse your answer to check whether your confidence will break or not
9. What are joins? Types of joins?
It was in two parts first is MCQ which contains 22 question mix of Logical Reasoning, Verbal Ability & Programming, and the second part was 2 coding questions. I have cleared this round and get call for interview .




1. Make in-place changes, that is, modify the nodes given a binary tree to get the required mirror tree.
The idea is to traverse recursively and swap the right and left subtrees after traversing the subtrees.
Follow the steps below to solve the problem:
Call Mirror for left-subtree i.e., Mirror(left-subtree)
Call Mirror for right-subtree i.e., Mirror(right-subtree)
Swap left and right subtrees.
temp = left-subtree
left-subtree = right-subtree
right-subtree = temp


Down: (row+1,col)
Right: (row, col+1)
Down right diagonal: (row+1, col+1)
This problem has the optimal substructure property. The path to reach (m, n) must be through one of the 3 cells: (m-1, n-1) or (m-1, n) or (m, n-1). So minimum cost to reach (m, n) can be written as “minimum of the 3 cells plus cost[m][n]”.
minCost(m, n) = min (minCost(m-1, n-1), minCost(m-1, n), minCost(m, n-1)) + cost[m][n]
Follow the below steps to solve the problem:
If N is less than zero or M is less than zero then return Integer Maximum(Base Case)
If M is equal to zero and N is equal to zero then return cost[M][N](Base Case)
Return cost[M][N] + minimum of (minCost(M-1, N-1), minCost(M-1, N), minCost(M, N-1))
The panel had Two members. one of them was from technical and another from the HR department.



'N' = 4,
4 can be represented as 2^2. So, 4 is the power of two, and hence true is our answer.
We can a^n (let’s say 3^5) as 3^4 * 3^0 * 3^1 = 3^5, so we can represent 5 as its binary i.e. 101
.calculate last bit(rightmost) bit of n
.if the last bit is 1 then multiply ans and a
.make an equal to the square of an as on every succeeding bit it got squared like a^0, a^1, a^2, a^4, a^8
1. Introduce yourself
2. What is your role in the company
3. Which language do you prefer python java
4 what is memory management in java
5 what is memory management in python
6 How garbage collector works in java
7 what is a cache
9 What are AI and cloud computing do you have any experience
10 print power of a given number without using multiple symbols in java
11 write SQL query to show 2nd topper of the class
12 what is a group by
13 Differences between truncate delete remove
14 Show polymorphism in java by writing code
15 real-life examples of Inheritance
HR Question
16 What you know about jio
17 Are you willing to reallocate
18 why do you want to leave this company
this round was for ignite and Illuminate after this round they declared the final result and I got selected as the illuminate round.
In this round, there were two interviewers 1st from HR and a second from a technical field



Method 2:
0) Make a board, and make a space to collect all solution states.
1) Start in the topmost row.
2) Make a recursive function that takes the state of the board and the current row number
as its parameter.
3) Fill a queen in a safe place and use this state of the board to advance to the next recursive
call, add 1 to the current row. Revert the state of the board after making the call.
a) Safe function checks the current column, left top diagonal, and right top diagonal.
b) If no queen is present then fill else return false and stop exploring that state
and track back to the next possible solution state
4) Keep calling the function till the current row is out of bound.
5) If the current row reaches the number of rows in the board then the board is filled.
6) Store the state and return.



Consider city 1 as the starting and ending point. Since the route is cyclic, we can consider any point as a starting point.
Generate all (n-1)! permutations of cities.
Calculate the cost of every permutation and keep track of the minimum cost permutation.
Return the permutation with minimum cost.
The interview started with basic questions
1. Introduction
2 Tell me about your internship
3 What stream API in java
4. Interface in java.
5. Is multiple inheritances possible, why is possible in c++ no in java
6. What is hibernate (Internship related question)
7. What is PostgreSQL
8 . Best sorting algorithm
9 Tell about merge sort
10 why merge sort better than quick sort
11 ask about n queen problem and traveling salesman problem then ask to write code
HR
the same person as 1st interview so does not ask me anything

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