Cracking the coding interview 6th edition pdf
Recycling
How to recognize
 
While there is a wide range of recurring problems, many similar recurring problems persist
patterns. A good indication that there is a recurring problem seems to be built with sub-problems.
 
When you hear a problem that  cracking the coding interview 6th edition pdf begins with the following, often (but not always)
 
Once again, practice makes perfect! The more problems you have, the easier it will be to identify you
recurring problems.
 
How to get closer

 
 
Recursive solutions, by definition, are solved from subproblems. Many times this
it will simply mean to calculate f (n) by adding, subtracting or otherwise
changing the solution to f (n-1). In other cases, you may need to do something more competitive
creased. However, we recommend the following approach:
 
1. Think about what the subproblem is. How many subproblems does f (n) depend on?
That is, in a recurring binary tree problem, each part is likely to be based on two problems
slogans. In a linked list problem, there will probably only be one.
 
2. Solutions for "lower case". That is, if you need to calculate f (n), first calculate it for f (0) or
f (1). This is usually a hard-coded value.
 
3. Solutions for f (2).
the precise process to translate the solutions of the subproblems into the actual solution.
 
5. Generalization for f (n).
 
"Bottom-up recycling" is often the simplest. Sometimes, however, it can be
useful for tackling "top-down" problems, where you just jump to a break
f (n) in its subproblems.
 
Things to keep in mind
 
1. All problems that can be solved recursively can be solved in another way (however
the code can be much more complex). Before diving into recursive code, ask
yourself how difficult it would be to implement this algorithm otherwise. Discuss the
exchange with your interviewer.
 
2. Recursive algorithms can be very spatially efficient. Each recurring call adds a new layer
for stacking, which means if your algorithm has recursive O (n) calls, you use it
Memory O (n). Oh! Here's a reason an iterative algorithm might be better.
orting and Searching
 
 
How to approach:
 
 
It is extremely valuable to understand common ranking algorithms, as many are ranking or
Search solutions require adjustments to known classification algorithms. Good focus when
you are given a question like this to run through the various sorting algorithms and see if
one of them is very good.
 
Example: You have a very large set of 'Person' objects. Sort the people in ascending order
years.
 
Here we are given two interesting facts: (1) It is a large set, so the efficiency is very good
important. (2) We sort by ages, so we know that the values ​​are in a small range. You
scanning the various classification algorithms, we might notice that the classification cube
Be a perfect candidate for this algorithm. In fact, we can make the small cubes (only 1 year
one) and find O (n) the current time.
 
Bubble sort:
 
 
Start at the beginning of an array and swap the first two elements if the first is greater than
the second. Go to the next pair etc, sweeping the array continuously until sorted.
0 (n A 2).
 
Selection order:
 
 
Find the smallest item using a linear scan and move it forward. So get the latest
it is smaller and smoother, doing a linear scan again. Continue doing this until all elements
instead. 0 (n A 2).
 
Combine classification:
 
 
Order each pair of items. Then order the four items by combining the two pairs. Then,
order the 8 elements, etc. 0 (n log n) worst and worst case scenario.
 
Quick sort:
 
 
Select a random characteristic and divide the matrix so that all the numbers are smaller than it.
They come in front of every element bigger than it. Then do that for each half, then each quarry
ter, etc. 0 (n log n) expected, 0 (n A 2) in the worst case.
 
Bucket type:
 
 
Divide the matrix into a limited number of deposits and then order each deposit individually.
This gives time 0 (n + m), where n is the number of elements and m is the given number
articles.