Tip 1: Never leave any topic from any chapter / Subject
Tip 2: Learn to explain your thoughts well.
Tip 3: Learn from previous experiences/interviews/problems asked.
Tip 4: At least four projects in Resume.
Tip 1: At least four projects on the resume
Tip 2: Do not write false things. You always get caught. Be genuine.
There were 2 DSA questions asked in this round.






INPUT : ARR [ ] = [ 1 , 2 , 3 , 4 , 5 ] , N = 4
OUTPUT: ARR [ ] = [ 1 , 2 , 4, 5 ]
The above example contains an odd number of elements, hence the middle element is clearly the (N+1) / 2th element, which is removed from the stack in the output.
INPUT : ARR [ ] = [ 5, 6, 7, 8 ] , N = 3
OUTPUT: ARR [ ] = [ 5, 7, 8 ]
The above example contains an even number of elements, so out of the two middle elements, we consider the one which occurs first. Hence, the middle element would be ((N+1) / 2 - 1) element, which is 6 and is removed from the stack in the output.
Initialize an empty stack temp and a variable count with 0.
Run a while loop till count becomes equal to half the initial size of the given stack
Pop the element of the given stack and push them in temp.
Pop the top element from the given stack.
Run a while loop till temp becomes empty.
Push the element of temp and push them in the given stack .
This round a lot of development questions were asked and 2 DSA questions were asked as well.



1. Quadruple p*q = r*s is the same as r*s = p*q.
2. If 2 or more products have the same count of quadruples, print the lowest value of the product i.e if (P1, P2) are the 2 products with the same count of such quadruples(C1 = C2) then 'P' = min(P1, P2).
3. If no such quadruple exists('C' = 0), return 0.
If the given array is [3, 4, 6, 2, 1], then the answer would be 6 1. Because there are two products 'P' i.e 6 and 12 which have the highest and same count 'C' of quadruples, i.e 'C' = 1. Therefore the lowest value of the product 'P' is the answer i.e 6.



For a (6 x 6) board, the numbers are written as follows:

You start from square 1 of the board (which is always in the last row and first column). On each square say 'X', you can throw a dice which can have six outcomes and you have total control over the outcome of dice throw and you want to find out the minimum number of throws required to reach the last cell.
Some of the squares contain Snakes and Ladders, and these are possibilities of a throw at square 'X':
You choose a destination square 'S' with number 'X+1', 'X+2', 'X+3', 'X+4', 'X+5', or 'X+6', provided this number is <= N*N.
If 'S' has a snake or ladder, you move to the destination of that snake or ladder. Otherwise, you move to S.
A board square on row 'i' and column 'j' has a "Snake or Ladder" if board[i][j] != -1. The destination of that snake or ladder is board[i][j].
You can only take a snake or ladder at most once per move: if the destination to a snake or ladder is the start of another snake or ladder, you do not continue moving - you have to ignore the snake or ladder present on that square.
For example, if the board is:
-1 1 -1
-1 -1 9
-1 4 -1
Let's say on the first move your destination square is 2 [at row 2, column 1], then you finish your first move at 4 [at row 1, column 2] because you do not continue moving to 9 [at row 0, column 0] by taking the ladder from 4.
A square can also have a Snake or Ladder which will end at the same cell.
For example, if the board is:
-1 3 -1
-1 5 -1
-1 -1 9
Here we can see Snake/Ladder on square 5 [at row 1, column 1] will end on the same square 5.

Here's your problem of the day
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What is recursion?