Tip 1: Concentrate on all the data structures and algorithms and practice at least 100 easy and 80 medium and 30-40 hard questions before the placements
Tip 2: Pick a coding language like 'Python' and practice well in that, so that during the hackathon we won't need to waste time exploring the language
Tip 1: As many projects as we have in the resume as good as it is.
Tip 2: Whatever is there in the resume we need to be strong in those areas. If we couldn't answer a question asked from the resume it would create a bad impression on us
This round was held around 4pm and it happened in online mode in hackerank platform. We were supposed to turn on our camera and mic during the whole round.



1. You can not slant the container i.e. the height of the water is equal to the minimum height of the two lines which define the container.
2. Do not print anything, you just need to return the area of the container with maximum water.

For the above Diagram, the first red marked line is formed between coordinates (2,0) and (2,10), and the second red-marked line is formed between coordinates (5,0) and (5,9). The area of water contained between these two lines is (height* width) = (5-2)* 9 = 27, which is the maximum area contained between any two lines present on the plane. So in this case, we will return 3* 9=27.
Since we assume there is a better answer in the sub-range of [i, j], then this optimal range must be contained by either [i + 1, j] or [i, j - 1], or both.
Let's assume [i + 1, j] doesn't contain the optimal range, but [i, j - 1] contains it. Then this means two things:
the optimal range is not in [i + 1, j - 1], otherwise, [i + 1, j] will contain it.
The optimal range contains the block [i, i + 1] (since this is the part which exists in [i, j - 1] but not in [i+1, j]).
However, notice that, len(i, j - 1) < len(i, j), and in the range [i, j], the area is constrained by the height of h[i] (even in the case h[i] == h[j]). Thus, in the range [i, j - 1], even all pillar are no shorter than h[j], the maximum area is smaller than the area formed by i & j, which contradicts our assumption there is a better answer in the sub-range of [i, j]. This contradiction suggests [i + 1, j] contains the optimal sub-range, if such sub-range exists.
According to above theorem, we can design the algorithm, whenever h[i] < h[j], we advance the pointer i.


Use matrix Breadth-first search to solve the problem
Hackathon
Regarding image processing
Converted image into 2D array and applied BFS and DFS to solve it

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