Tip 1 : Clear all the topics related to Android with deep details about every topics.
Tip 2 : Practice DSA and algorithms from GFG, CodeStudio daily.
Tip 3 : Dry run code before running the code.
Tip 1 : Keep it to the point don't add anything for the sake of making it large.
Tip 2 : Put atleast 2 good projects if you are a fresher.



Minimum operations to make ‘STR’ ‘0010’ beautiful is ‘1’. In one operation, we can convert ‘0’ at index ‘0’ (0-based indexing) to ‘1’. The ‘STR’ now becomes ‘1010’ which is a beautiful string.



1. If you encounter a situation when 'B[i]' is greater than the number of remaining nodes in the list, then simply reverse the remaining nodes as a block and ignore all the block sizes from 'B[i]'.
2. All block sizes are contiguous i.e. suppose that block 'B[i]' ends at a node cur, then the block 'B[i+1]' starts from the node just after the node cur.
Linked list: 1->2->3->4->5
Array B: 3 3 5
Output: 3->2->1->5->4
We reverse the first block of size 3 and then move to block 2. Now, since the number of nodes remaining in the list (2) is less than the block size (3), we reverse the remaining nodes (4 and 5) as a block and ignore all the block sizes that follow.



Consider following matrix:

The rectangle (1,1) to (3,3) is the rectangle with the maximum sum, i.e. 29.

Difference between constraint and relative layout Kotlin co routines.
Why kotlin over java.
Fragment life cycle with a live app explanation.
How to import 3d models in an app Broadcast receivers and services.
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What is recursion?