Tip 1: Prepare DSA questions thoroughly.
Tip 2: Be confident during the interview.
Tip 3: Prepare your project very well and be able to explain it clearly.
Tip 1: Add links to your coding profiles in resume.
Tip 2: Add strong, impactful projects to your resume that demonstrate practical application of your skills.
In the OA, here were 20 MCQs on OOPS, DBMS and 2 coding problems.



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.
1:- Initialize the DP array.
2:- Iterate through each coin.
3:- Update the number of ways for each possible sum using the current coin.
4:- Return the final number of ways.



You may assume that duplicates do not exist in the given traversals.
For the preorder sequence = [1, 2, 4, 7, 3] and the inorder sequence = [4, 2, 7, 1, 3], we get the following binary tree.

1:- Pick the root node from the first element of preorder.
2:- Find the root’s index in inorder.
3:- Split inorder into left and right subtrees.
4:- Recursively build left and right subtrees.
5:- Perform postorder traversal to get the result.



• The left subtree of a node contains only nodes with data less than the node’s data.
• The right subtree of a node contains only nodes with data greater than the node’s data.
• Both the left and right subtrees must also be binary search trees.
'P' = 1, 'Q' = 3
tree = 2 1 4 -1 -1 3 -1 -1 -1,
The BST corresponding will be-

Here, we can clearly see that LCA of node 1 and node 3 is 2.
1:- Start at the root.
2:- If both p and q values < root value, move to root.left.
3:- If both p and q values > root value, move to root.right.
4:- Otherwise (they split or one equals root) the current root is the LCA — return it.


The width of each bar is the same and is equal to 1.
Input: ‘n’ = 6, ‘arr’ = [3, 0, 0, 2, 0, 4].
Output: 10
Explanation: Refer to the image for better comprehension:

You don't need to print anything. It has already been taken care of. Just implement the given function.
1:- Initialize two arrays arr1 and arr2 of size n.
2:- Fill arr1[i] with the maximum height from start up to index i.
3:- Fill arr2 [i] with the maximum height from end up to index i.
4:- For each index i, compute water[i] = min(arr1[i], arr2 [i]) - height[i].
5:- Sum up all water[i] values to get the total trapped water.

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