## Algorithms: Why is $O (| V | + | E |)$ linked to DFS and BFS instead of just $O (| E |)$?

In a sense, I understand why the limit on BFS and DFS is $$O (| V | + | E |)$$. Each vertex is considered, hence the $$O (| V |)$$, and in the course of considering all adjacent vertices, we end up considering $$2 | E | = O (| E |)$$ total vertices because $$sum_ {v in V} d_v = 2 | E |$$. What I don't understand is why we bother to specify $$O (| V | + | E |)$$ first of all because it seems to me that the tightest limit you can really have in any situation is $$O (| E |)$$. I mean, it seems to me that the $$O (| V |)$$ It doesn't help you at all. To see why, consider two cases:

• The graphic is connected. Then of course $$| E | ge | V – 1 |$$.
• The graphic is not connected. Then you are "stuck" in the connected component where you start. Again, within the connected component $$A$$ you start you have $$| E_A | ge | V_A – 1 |$$. Perhaps there are more edges and vertices in other places. But once again the $$O (| V |)$$ It seems redundant.

Another way I think about it: adding a vertex alone doesn't create more work since the vertex will simply be floating in space. Add borders can create extra work.

So why / when is the $$O (| V |)$$ Useful?

## complexity theory: why are upstream selector matching algorithms more efficient?

Most implementations of CSS-like selectors (that is, patterns that can match paths in a tree) seem to use an upward approach. That is, for each node in the tree, they verify a match with the rightmost element of the selector pattern. If the node matches, they look for their ancestors looking for a match with the element of the rightmost selector pattern, and so on.

My question is, why is this more efficient than starting from above and working down (from left to right in the selector pattern)? In both approaches, it seems that you have to examine each route, and both approaches can benefit from the short circuit.

## Algorithms: string counting with balanced substrings

For a good rope $$s$$ of length $$k$$, leave $$M (n, s)$$ denote the number of length chains $$n$$ ending in $$s$$ in which each $$k$$-the newsletter substring is good. The amount $$M (n, s)$$ It is given by the following recurrence:
begin {align} & M (k, s) = 1 \ & M (n + 1, ta) = M (n + 1, tb) = M (n, at) + M (n, bt) \ & M (n + 1, tc) = M (n, ct) end {align}
here $$t$$ it's a length chain $$k-1$$. Summing up all the good chains $$s$$, we can solve your problem

As $$to$$ Y $$b$$ are really interchangeable, we can also consider a slightly different recurrence, in which $$s$$ it only consists of $$to$$& # 39; s and $$c$$& # 39; s:
begin {align *} & M & # 39; (k, s) = 2 ^ { #_ a (s)} \ & M & # 39; (n + 1, ta) = 2M & # 39; (n, en) \ & M & # 39; (n + 1, tc) = M & # 39; (n, ct) end {align *}
Here the second parameter is a "template" in which $$to$$ means any $$to$$ or $$b$$.

Using any of the recurrences, you can find explicit formulas for any given value of $$k$$.

## Advanced SEO techniques for success in 2020

In recent years, SEO has changed a lot. In its infancy, SEO was limited to filling in keywords, spam links and adjusting the background code. Fortunately, Google has eradicated many practices that were only designed to manipulate search results.

To improve the quality of search engine results, Google continues to modernize its algorithms. As a result, the SEO techniques that were used in the first decade of the 21st century are no longer effective today.

For that reason, online companies must keep up with the latest search engine algorithms and update their SEO techniques over time.

In the next section, we will talk about some SEO techniques that will help you generate leads in 2020 and the coming years. Then, without further delay, let's start.

.

## Matrices – Potential algorithms to solve coding problems?

so I just tried a technical test online. I was in the hacker range, however, I am sure that leetcode generally has the same (and more), so I thought about asking.

I can't think of a good solution and I can't find a leet code exercise that matches what I was asked for, so I'm here for someone to point me in the right direction.

The problem:

Someone has to carry water in cans and can only carry one can at a time. I was given 2 matrices, one of which had the total amount of water that each can can hold and the second has the amount of water that each can used carefully. I had to find the minimum amount of cans needed to transport all the water used.

An example of input was:

Total = (3, 5, 3, 5, 5)

Used = (3, 2, 1, 3, 1)

So can (0) has a total of 3 and is currently full. The can (1) has a total of 5, but only 2, etc., etc. are being used. I had to find the minimum amount of cans needed to transport all the water.
The solution in this example was 2 since you can take the water from the can (0) and put it in the can (1) to fill it. Then, you can drink water from the cans (2) and (3) and put them in the can (4), therefore, you only need cans (2) and (4) to transport all the water.

Does anyone recognize this question and could I address the problem in the leet or hackerrank code since I can't find a good algorithm myself?

## algorithms: criteria based on the due date of a task

Basically we have a ticket system, every day we receive multiple tickets with different categorization (Question, Change, Error) and importance (High, Medium, Low) for different customers (A, B, C). My job is to create some type of system that determines the due date of these "tasks" based on

• The status of the above, completed or not.
• The availability of developers.
• A criterion of the attributes mentioned above (categorization, importance and customer
kind).

How can that be achieved?

## What is the history of Google algorithms …?

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What is the history of Google algorithms …?

1. ## What is the history of Google algorithms …?

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## What is the history of Google algorithms …?

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What is the history of Google algorithms …?

## Algorithms – What is the great complexity of time $O$ of this nested loop for

I want the complexity of time to be great, oh $$Or$$ as well as the value of the count in terms of $$n$$

From my understanding,
the first for the cycle runs for $$lg (n)$$ times where $$lg (x) = log_2 (x)$$
Also, every time, $$i$$ it's the way $$2 ^ k$$ where $$; k en N cup {0 }$$ Y $$j$$ run from $$1$$ to $$i-1 = 2 ^ k -1 ; forall k en N cup {0 },$$ such that $$2 ^ k

Thus,
$$account = sum_ {k = 0} ^ {t} (2 ^ k – 1)$$ where $$2 ^ t geq n rightarrow t geq lg (n)$$. For the closest approximation to the count, take $$t = lg (n) ; text {{does this make sense?}}$$ Then we get:

$$count = sum_ {k = 0} ^ { lg (n)} (2 ^ {k} -1) = frac {1 times (2 ^ { lg (n)} – 1)} { 2-1} – ( lg (n) +1) = n- lg (n) -2$$. So is this $$O (n)$$?

## Algorithms: What problems does a university / college have that can be solved in the form of software development?

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