Tip 1 : You should have in depth knowledge of your projects. Sometimes Adobe spend a whole interview in project discussion (in some cases).
Tip 2 : OOPS is a very important subject if you are preparing for Adobe. Also one should know real life examples of it and should be able to code any concept of it.
Tip 3 : One should have good knowledge of data structures. Mainly Array, Math, Tree, Recursion and LinkedList.
Tip 4 : According to me , projects just play role if you applying off campus and that too for the shortlisting of your resume as it gives you an edge in respect to other candidates. So if you are applying off campus you should have atleast 2-3 good projects in you resume.
Tip 1 : One should have good projects.
Tip 2 : The presentation of your resume should be really good. Bold keywords like tech stack you used or the topics you are really good at.
I just run a loop from 1 till N
Step 1 : While n is divisible by 2, print 2 and divide n by 2.
Step 2 : After step 1, n must be odd. Now start a loop from i = 3 to square root of n. While i divides n, print i and divide n by i. After i fails to divide n, increment i by 2 and continue.
Step 3 : If n is a prime number and is greater than 2, then n will not become 1 by above two steps. So print n if it is greater tha...
I wrote a solution using Dynamic Programming.
Step 1 : initialize a 2 D matrix and assign a meaning to value (here in our case , possible paths from given point to bottom right corner)
Step 2 : I initialized the matrix using base cases (if one column or one row)
Step 3 : filled the matrix accordingly.
I used Dynamic programing.
Step 1 : initialized a 1-D array
Step 2 : initialized last index value to zero.
Step 3 : calculated minimum steps from n-2 index to n-1 index and till index = 0
Step 1 : I used something similar to sliding window algorithm
Step 2 : I called a function which contains starting index of the string , an empty StringBuilder as it is mutable and an Arraylist to store indexes of all valid substrings as parameters.
Step 3 : declared a base case which states that the whole string is traversed.
Step 4 : declared some conditions,
(a) if the output Str...
I kept an counter to calculate the level and and performed level order traversal and applied a condition that if the level was odd I added the elements into the list in reverse order fashion.