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I am struggling optimising this past amazon Interview question involving a DAG.
This is what I tried (The code is long and I would rather explain it)-
Basically since the graph is a DAG and because its a transitive relation a simple traversal for every node should be enough.
So for every node I would by transitivity traverse through all the possibilities to get the end vertices and then compare these end vertices to get
the most noisy person.
In my second step I have actually found one such (maybe the only one) most noisy person for all the vertices of the traversal in step 2. So I memoize all of this in a mapping and mark the vertices of the traversal as visited.
So I am basically maintaining an adjacency list for the graph, A visited/non visited mapping and a mapping for the output (the most noisy person for every vertex).
In this way by the time I get a query I would not have to recompute anything (in case of duplicate queries).
The above code works but since I cannot test is with testcases it may/may not pass the time limit. Is there a faster solution(maybe using DP) to this. I feel I am not exploiting the transitive and anti symmetric condition enough.
Obviously I am not checking the cases where a person is less wealthy than the current person. But for instance if I have pairs like - (1,2)(1,3)(1,4)...etc and maybe (2,6)(2,7)(7,8),etc then if I am given to find a more wealthy person than 1 I have traverse through every neighbor of 1 and then the neighbor of every neighbor also I guess. This is done only once as I store the results.
Question Part 1
Question Part 2
Edit(Added question Text)-
Rounaq is graduating this year. And he is going to be rich. Very rich. So rich that he has decided to have
a structured way to measure his richness. Hence he goes around town asking people about their wealth,
and notes down that information.
Rounaq notes down the pair (Xi; Yi) if person Xi has more wealth than person Yi. He also notes down
the degree of quietness, Ki, of each person. Rounaq believes that noisy persons are a nuisance. Hence, for
each of his friends Ai, he wants to determine the most noisy(least quiet) person among those who have
wealth more than Ai.
Note that "has more wealth than"is a transitive and anti-symmetric relation. Hence if a has more wealth
than b, and b has more wealth than c then a has more wealth than c. Moreover, if a has more wealth than
b, then b cannot have more wealth than a.
Your task in this problem is to help Rounaq determine the most noisy person among the people having
more wealth for each of his friends ai, given the information Rounaq has collected from the town.
Input
First line contains T: The number of test cases
Each Test case has the following format:
N
K1 K2 K3 K4 : : : Kn
M
X1 Y1
X2 Y2
. . .
. . .
XM YM
Q
A1
A2
. . .
. . .
AQ
N: The number of people in town
M: Number of pairs for which Rounaq has been able to obtain the wealth
information
Q: Number of Rounaq’s Friends
Ki: Degree of quietness of the person i
Xi; Yi: The pairs Rounaq has noted down (Pair of distinct values)
Ai: Rounaq’s ith friend
For each of Rounaq’s friends print a single integer - the degree of quietness of the most noisy person as required or -1 if there is no wealthier person for that friend.
Perform a topological sort on the pairs X, Y. Then iterate from the most wealthy down the the least wealthy, and store the most noisy person seen so far:
less wealthy -> most wealthy
<- person with lowest K so far <-
Then for each query, binary search the first person with greater wealth than the friend. The value we stored is the most noisy person with greater wealth than the friend.
UPDATE
It seems that we cannot rely on the data allowing for a complete topological sort. In this case, traverse sections of the graph that lead from known greatest to least wealth, storing for each person visited the most noisy person seen so far. The example you provided might look something like:
3 - 5
/ |
1 - 2 |
/ |
4 --
Traversals:
1 <- 3 <- 5
1 <- 2
4 <- 2
4 <- 5
(Input)
2 1
2 4
3 1
5 3
5 4
8 2 16 26 16
(Queries and solution)
3 4 3 5 5
16 2 16 -1 -1
I am building a website where I need to make sure that the number of "coins" and number of "users" wont kill the database if increases too quickly. I first posted this on mathematica (thinking its a maths website, but found it it's not). If this is the wrong place, please let me know and I'll move it accordingly. However, it does boil down to solving a complex problem: will my database explode if the users increase too quickly?
Here's the problem:
I am trying to confirm if the following equations would work for my problem. The problem is that i have USERS (u) and i have COINS (c).
There are millions of different coins.
One user may have the same coin another user has. (i.e. both users have coin A)
Users can trade coins with each other. (i.e. Trade coin A for coin B)
Each user can trade any coin with another coin, so long as:
they don't trade a coin for the same coin (i.e. can't trade coin A for another coin A)
they can't trade with themselves (i.e. I can't offer my own Coin A for my own Coin B)
So, effectively, there are database rows stored in the DB:
trade_id | user_id | offer_id | want_id
1 | 1 | A | B
2 | 2 | B | C
So in the above data structure, user 1 wants coin A for coint B, and user 2 wants coin B for coin C. This is how I propose to store the data, and I need to know that if I get 1000 users, and each of them have 15 coins, how many relationships will get built in this table if each user offers each coin to another user. Will it explode exponentially? Will it be scalable? etc?
In the case of 2 users with 2 coins, you'd have user 1 being able to trade his two coins with the other users two coins, and vice versa. That makes it 4 total possible trade relationships that can be set up. However, keeping in mind that if user 1 offers A for B... user 2 can't offer B for A (because that relationship already exists.
What would the equation be to figure out how many TRADES can happen with U users and C coins?
Currently, I have one of two solutions, but neither seem to be 100% right. The two possible equations I have so far:
U! x C!
C x C x (U-1) x U
(where C = coins, and U = users);
Any thoughts on getting a more exact equation? How can I know without a shadow of a doubt, that if we scale to 1000 users with 10 coins each, that this table won't explode into millions of records?
If we just think about how many users can trade with other users. You could make a table with the allowable combinations.
user 1
1 | 2 | 3 | 4 | 5 | 6 | ...
________________________________
1 | N | Y | Y | Y | Y | Y | ...
user 2 2 | Y | N | Y | Y | Y | Y | ...
3 | Y | Y | N | Y | Y | Y | ...
The total number of entries in the table is U * U, and there are U N's down the diagonal.
Theres two possibilities depending on if order matters. Is trade(user_A,user_B) is the same as trade(user_B,user_A) or not? If order matters the same the number of possible trades is the number of Y's in the table which is U * U - U or (U-1) * U. If the order is irrelevant then its half that number (U-1) * U / 2 which are the Triangular numbers. Lets assume order is irrelevant.
Now if we have two users the situation with coins is similar. Order does matter here so it is C * (C-1) possible trades between the users.
Finally multiply the two together (U-1) * U * C * (C-1) / 2.
The good thing is that this is a polynomial roughly U^2 * C^2 so it will not grow to quickly. This thing to watch out for is if you have exponential growth, like calculating moves in chess. Your well clear of this.
One of the possibilities in your question had U! which is the number of ways to arrange U distinct objects into a sequence. This would have exponential growth.
There are U possible users and there are C possible coins.
Hence there are OWNS = CxU possible "coins owned by an individual".
Hence there are also OWNS "possible offerings for a trade".
But a trade is a pair of two such offerings, restricted by the rule that the two persons acting as offerer cannot be the same, and neither can the offered coin be the same. So the number of candidates for completing a "possible offering" to form a "complete trade" is (C-1)x(U-1).
The number of possible ordered pairs that form a "full-blown trade" is thus
CxUx(C-1)x(U-1)
And then this is still to be divided by two because of the permutation issue (trades are a set of two (person,coin) pairs, not an ordered pair).
But pls note that this sort of question is actually an extremely silly one to worry about in the world of "real" database design !
I need to know that if I get 1,000 users, and each of them have 15 coins, how many relationships will get built in this table if each user offers each coin to another user.
The most that can happen is all 1,000 users each trade all of their 15 coins, for 7,500 trades. This is 15,000 coins up for trade (1,000 users x 15 coins). Since it takes at least 2 coins to trade, you divide 15,000 by 2 to get the maximum number of trades, 7,500.
Your trade table is basically a Cartesian product of the number of users times the number of coins, divided by 2.
(U x C) / 2
I'm assuming users aren't trading for the sake of trading. That they want particular coins and once they get the coins, won't trade again.
Also, most relational databases can handle millions and even billions of rows in a table.
Just make sure you have an index on trade id and on user id, trade id in your Trade table.
The way I understand this is that you are designing an offer table. I.e. user A may offer coin a in exchange for coin b, but not to a specific user. Any other user may take the offer. If this is the case, the maximum number of offers is proportional to the number of users U and the square of the number of coins C.
The maximum number of possible trades (disregarding direction) is
C(C-1)/2.
Every user can offer all the possible trades, as long as every user is offering the trades in the same direction, without any trade being matched. So the absolute maximum number of records in the offer table is
C(C-1)/2*U.
If trades are allowed between more than two user the number decreases above half that though. E.g. if A offers a for b, B offers b for c and C offers c for a. Then a trade could be accomplished in a triangle by A getting b from B, B getting c from C and C getting a from A.
The maximum number of rows in the table can then be calculated by splitting the C coins into two groups and offering any coin it the first group in exchange any coin in the second. We get the maximum number of combination if the groups are of the same size, C/2. The number of combinations is
C/2*C/2 = C^2/4.
Every user may offer all these trades without there being any possible trade. So the maximum number of rows is
C^2/4*U
which is just over half of
C(C-1)/2*U = 2*(C^2/4*U) - C/2*U.
I have created XMLA to process cube in incremental way. It is using type "ProcessUpdate" for dimensions and "ProcessAdd" for measurement partitions. I am facing one issue on distinct count. Let me take one example:
Order Id CustId Amount
1 C1 100.00
2 C2 200.00
3 C3 300.00
4 C4 400.00
5 C5 500.00
If we browse cube, SSAS shows sum of orders as 1500.00, and distinct customer count for all orders as 5. Now adding new fact record for cancelling one order,
e.g.:
Order Id CustId Amount
3 C3 -300.00
After incremental processing, it shows sum of orders as 1200.00 which is correct one. But the distinct customer count for all orders keeps same and shows 5 which is incorrect.
I can understand that the rows are getting append on incremental process which works for sum operation, but fails computing distinct count. I want to know if there is any way that can remove order #3 from all aggregate operation while processing in incremental way.
Distinct customer count remaining at five is correct, as it doesn't know that a minus 300 means the customer shouldn't show. If you processed the cube fully it would show as 5 distinct customers.
This isn't to do with incremental processing, this is to do with how SSAS handles distinct count - It's just "Count distinct customer Ids in fact table", and C3 is there twice with a sale of 300 and a sale of -300.
You need to reconsider how to handle this, ideally at the stage where you load your data warehouse. You could handle it in MDX by not including anyone who has sales of zero or less, but then the whole distinct count calculation will be done in MDX and will be much, much slower.
I have two separate vectors HG and AG which represent two different soccer scores. One of them is the amount of home goals while the other is the amount of away goals. I'd like to know if there is a way of counting how many times a result occurs and putting it into a table, e.g. if value is 1 from HG and 2 from AG then the result is 1-2. Then I would like to find out how many times this score occurs.
vector<-paste(HG,AG,sep="-")
result<-data.frame(table(vector))
This is more of a maths question than programming but I figure a lot of people here are pretty good at maths! :)
My question is: Given a 9 x 9 grid (81 cells) that must contain the numbers 1 to 9 each exactly 9 times, how many different grids can be produced. The order of the numbers doesn't matter, for example the first row could contain nine 1's etc. This is related to Sudoku and we know the number of valid Sudoku grids is 6.67×10^21, so since my problem isn't constrained like Sudoku by having to have each of the 9 numbers in each row, column and box then the answer should be greater than 6.67×10^21.
My first thought was that the answer is 81! however on further reflection this assumes that the 81 numbers possible for each cell are different, distinct number. They are not, there are 81 possible numbers for each cell but only 9 possible different numbers.
My next thought was then that each of the cells in the first row can be any number between 1 and 9. If by chance the first row happened to be all the same number, say all 1s, then each cell in the second row could only have 8 possibilites, 2-9. If this continued down until the last row then number of different permutations could be calculated by 9^2 * 8^2 * 7^2 ..... * 1^2. However this doesn't work if each row doesn't contain 9 of the same number.
It's been quite a while since I studied this stuff and I can't think of a way to work it out, I'd appreciate any help anyone can offer.
Imagine taking 81 blank slips of paper and writing a number from 1 to 9 on each slip (nine of each number). Shuffle the deck, and start placing the slips on the 9x9 grid.
You'd be able to create 81! different patterns if you considered each slip to be unique.
But instead you want to consider all the 1's to be equivalent.
For any particular configuration, how many times will that configuration be repeated
due to the 1's all being equivalent? The answer is 9!, the number of ways you can permute the nine slips with 1 written on them.
So that cuts the total number of permutations down to 81!/9!. (You divide by the number of indistinguishable permutations. Instead of 9! indistinguishable permutations, imagine there were just 2 indistinguishable permutations. You would divide the count by 2, right? So the rule is, you divide by the number of indistinguishable permutations.)
Ah, but you also want the 2's to be equivalent, and the 3's, and so forth.
By the same reasoning, that cuts down the number of permutations to
81!/(9!)^9
By Stirling's approximation, that is roughly 5.8 * 10^70.
First, let's start with 81 numbers, 1 through 81. The number of permutations for that is 81P81, or 81!. Simple enough.
However, we have nine 1s, which can be arranged in 9! indistinguishable permutations. Same with 2, 3, etc.
So what we have is the total number of board permutations divided by all the indistinguishable permutations of all numbers, or 81! / (9! ** 9).
>>> reduce(operator.mul, range(1,82))/(reduce(operator.mul, range(1, 10))**9)
53130688706387569792052442448845648519471103327391407016237760000000000L