Tip 1 : Practice medium level problems.
Tip 2 : Brush up computer fundamentals from subjects like OS, DBMS and CN.
Tip 3 : Have a good project or good internship experience and have in-depth knowledge regarding what you have done.
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It was day time. Interviewer made a comfortable environment



Let the array 'ARR' be [1, 2, 3] and 'B' = 5. Then all possible valid combinations are-
(1, 1, 1, 1, 1)
(1, 1, 1, 2)
(1, 1, 3)
(1, 2, 2)
(2, 3)
Step 1 : I first used 3 loops.
Step 2 : Interviewer asked me to optimize the solution.
Step 3 : Then I gave solution with sorting the array and using 2 pointer approach.



For Amount = 70, the minimum number of coins required is 2 i.e an Rs. 50 coin and a Rs. 20 coin.
It is always possible to find the minimum number of coins for the given amount. So, the answer will always exist.
Step 1 : I first used for loops. It was not good enough.
Step 2 : Interviewer asked me to optimize the solution.
Step 3 : Then I gave dp solution. He asked me to create DP table and explain every value and explain how I reached there.
2 DSA questions were asked.


1. The grid has 0-based indexing.
2. A rotten orange can affect the adjacent oranges 4 directionally i.e. Up, Down, Left, Right.
I gave BFS solution and explained every part .



Input : 1 -> 2 -> 3 -> 4 -> 'NULL' and 'K' = 2
Output: 1 -> 2 -> 4 -> 'NULL'
Explanation:
After removing the second node from the end, the linked list become 1 -> 2 -> 4 -> 'NULL'.

Explained the approach by using 2 traversals - First to find the length of list and then to delete the node.
Also, discussed an approach to do it in one traversal.



Consider the two strings 'P' = "abfyg" and 'Q' = "gabfy"
If we cyclically rotate String 'P' to the right once. The resulting string P becomes "gabfy" which is equal to String 'Q'.
Therefore it is possible to convert String 'P' to String 'Q'.
Concatenate the string twice and check if the string is a substring of the concatenated one.



1. You are not required to print the output explicitly, it has already been taken care of. Just implement the function and return the minimum number of parentheses required to make a string valid.
Use stacks and insert opening parenthesis until it reaches the desired number. Also insert the number to know the number of parentheses already there in the stack. And if there is a number that is smaller than the top of stack, pop elements from the stack until it satisfies the number.

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