Tip 1 - Practice Atleast 250 Questions from geeks from geeks and coding ninjas
Tip 2 - Ex- Do atleast 2 good projects
Tip 1 : Have some projects on resume.
Tip 2 : Do not put false things on resume and be confident.
2 coding questions



Let S = “abdd” and X = “bd”.
The windows in S which contain all the characters in X are: 'abdd', 'abd', 'bdd', 'bd'.
Out of these, the smallest substring in S which contains all the characters present in X is 'bd'.
All the other substring have a length larger than 'bd'.
1- Generate all substrings of string1 (“this is a test string”)
2- For each substring, check whether the substring contains all characters of string2 (“tist”)
3- Finally, print the smallest substring containing all characters of string2.






You can only stack a box on top of another box if the dimensions of the 2-D base of the lower box ( both length and width ) are strictly larger than those of the 2-D base of the higher box.
You can rotate a box so that any side functions as its base. It is also allowed to use multiple instances of the same type of box. This means, a single type of box when rotated, will generate multiple boxes with different dimensions, which may also be included in stack building.

The height, Width, Length of the type of box will interchange after rotation.
No two boxes will have all three dimensions the same.
Don’t print anything, just return the height of the highest possible stack that can be formed.
) Generate all 3 rotations of all boxes. The size of rotation array becomes 3 times the size of the original array. For simplicity, we consider width as always smaller than or equal to depth.
2) Sort the above generated 3n boxes in decreasing order of base area.
3) After sorting the boxes, the problem is same as LIS with following optimal substructure property.
MSH(i) = Maximum possible Stack Height with box i at top of stack
MSH(i) = { Max ( MSH(j) ) + height(i) } where j < i and width(j) > width(i) and depth(j) > depth(i).
If there is no such j then MSH(i) = height(i)
4) To get overall maximum height, we return max(MSH(i)) where 0 < i < n



A 'deep copy' of a linked list means we do not copy the references of the nodes of the original linked list, rather for each node in the original linked list, a new node is created.
1) Create all nodes in copy linked list using next pointers.
2) Store the node and its next pointer mappings of original linked list.
3) Change next pointer of all nodes in original linked list to point to the corresponding node in copy linked list.
This was an HR round with the CEO which went on for about 25 mins. For me this was the most fun round. Firstly he asked normal questions like, describe a tough situation you have faced, describe your most interesting project etc. When he got to know I play the guitar, he asked me a design question – how would you detect the raga in a song, which led to a less-technical and more music based discussion. In the end he asked why I wanted to join CodeNation. This interview ended with me asking him 2 questions about CodeNation.
Firstly he asked normal questions like, describe a tough situation you have faced, describe your most interesting project etc. When he got to know I play the guitar, he asked me a design question – how would you detect the raga in a song, which led to a less-technical and more music based discussion. In the end he asked why I wanted to join CodeNation. This interview ended with me asking him 2 questions about CodeNation.
Believe in yourself.
Keep calm and get placed.

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What is recursion?