I have a problem, input will contain loggedin user id and timestamp when user logged in. when user logs in again need to find out no of times user logged in last x seconds.
input =
{(P1, 0),
(P2, 1),
(P3, 2),
(P1,3),
(P1,4),
(P2,5),
(P1,6)}
Q1: For last 4 seconds for P1 need to output how many times user has logged in.
output: P1,2
Q2: In last 6 seconds for P1:
output: P1,4
To solve this, I have initially used a hashMap, with keys as person id and values as set of timestamps for each person.
Map<personID,set<TimeStamp>> personMap = new HashMap();
personMap.put(P1,Treeset(0,3,4));
personMap.put(P2,Treeset(1,5));
personMap.put(P3,Treeset(2));
then on the final entry, we retrive the treeset and find values that fall after last 4 seconds from given time. This is working, but I am confused on timecomplexity, what will be timecomplexity for inserting and retrival considering values are Treesets?
Is there any better datastructure of storing timestamps in personMap?
The bottom layer of Treeset is red-black tree, and the time complexity of query is O (log n),
Compared with HashSet, TreeSet has slightly lower performance. The time complexity of insert query and other operations is O (log n).
HashMap is recommended.
Time Complexity:
tldr: I'm pretty sure this takes O(n) per query and O(log(n)) per insert.
For each query, you initially use log(n) on the TreeSet to select the right user.
Once you have the user, it sounds like you are finding the first timestamp within your bound (again can be log(k)) and iterating through the rest in your bound. This part becomes O(k) because you could query for a very long time back and end up iterating through your whole set.
Assuming your queries are random, the time complexity looks something like O(log(N) + K/2 + log(K)) -> O(K/2) -> O(n) (K is the set size for the user size, N is user count).
Solution:
tldr: Use a prefix sum.
The one way I can think of optimizing this from O(K) into O(log(K)) would be to keep track of the total number of logins performed by the user and store that with each timestamp.
Something like:
P1#Time0: 1 total login
P1#Time5: 2 total logins
P1#Time7: 3 total logins
Then for each query, you can just query the first time stamp out of your range and subtract its login count with the current login count.
Example: You login at time = 9 and query for all logins within 4 seconds prior.
You have:
P1#Time0: 1 total login
P1#Time5: 2 total logins
P1#Time7: 3 total logins
P1#Time9: 4 total logins (just now added)
Query finds Time0 as upper bound in log(n):
P1#Time0: 1 total login <- closest time outside 4 seconds
P1#Time5: 2 total logins
P1#Time7: 3 total logins
P1#Time9: 4 total logins
Take 4 - 1 (from their login counts) = 3, you have logged in three times from your current login to 4 seconds ago (this method counts your current login as well, subtract one from your result if you don't want that).
Look up a prefix sum for more info on this.
One note about this, it assumes that any newly added entries will come later than previous ones or you have the entire set of timestamps at the beginning. Your question sounds a lot like one coming from leetcode and if that's the case they most likely would put a condition like this in there. Otherwise, the solution might not function for what you need.
Related
I have a project that requires me to search though a list given a name time and status, and write a loop that loop that initializes at 0 and begins to count the number of occurrences when the number variable changes from X to Y.
For example take 2 apps with the given data
application:xyz time(min):0 status:200
application:abc time(min):0 status:404
application:xyz time(min):1 status:404
application:abc time(min):1 status:200
application:xyz time(min):2 status:404
application:abc time(min):2 status:404
the calculation would go like Looking at XYZ, it starts out in 200 but changes to 404( which we will call N). ABC transitions to 200 in minute 1, transitions to 404 in minute 2. both have a total runtime of 2 minutes (S).
So for both XYZ and ABC we get
N=1 and S=2 and Ans=S/N =2 minutes
I cant seem to wrap my head around the logic because i am supposed to be getting the data from a read csv. and then do this calculation. Any tips on the method?
There are n vendors on amazon who are selling the product at particular price at particular time, I have to design an algorithm which will select the product with least price at particular time.
For ex: For below set of input
Input Format :
<StartTime, EndTime, Price of product by this vendor in this time frame>
1 5 20
3 8 15
7 10 8
Output should be:
1 2 20
3 6 15
7 10 8
I have done with the solution by storing the prices corresponding to time in hashmap, and updating the price if there exist a price lesser then the old one corresponding to that time, and then made the list in vendor class to store all the times corresponding to the particular price.
But above solution is taking O(n2) time complexity, so searching for some fancy DS or approach to solve this in lesser time complexity.
You can use a sweep line algorithm and a multiset to solve it in O(N log N) time:
Let's create two events for each vendor: the moment she starts selling the item and the moment she ends. We'll also create one "check" event for each time we're interested in.
Now we'll sort the list of events by their times.
For each event, we do the following: if it's a start event, we add the new price to the multiset. Otherwise, we remove it.
At any moment of time, the answer is the smallest element in the multiset, so we can answer each query efficiently.
If the multiset supports "fast" (that is, O(log N) or better) insertions, deletions and finding the smallest element, this solution uses O(N log N) time and O(N) space. There is no mulitset in the Java standard library, but you can use a TreeSet of pairs (price, vendor_id) to work around this limitation.
Recently I have developed and IMPLEMENTED an algorithm which is performing following 4 operations in Non-Amortized O(log m) time , where m is the number of elements present in Memory=O(m).
1.>insert(int index,int data)::
It inserts data at any index(just like in array) randomly and dynamically but with different time complexity.It can be understood as inserting in an array at a particular index but main feature is that it can insert data at same index in O(log m) time along with SHIFTING ALL data present contiguously from that index onwards .For eg:: insert any data as:
index ,data
{
(1 ,0),
(2 ,1),
(78 ,2),
(0 ,3),
(45 ,4),
(58999,5),
(32111,6),
(1 ,7),
(78 ,8),
(78 ,9),
(78 ,-1),
(0 ,-2),
(0 ,-3),
(0 ,-4),
(23 ,-5)
}. Then total time complexity for all insertions is O(log (m!) ), here m=15 and Memory= O(15).
NOTE:: NEW SEQUENCE OF DATA IS STORED as per my ALGORITHM AS::
{
(0,-4),
(1,-3),
(2,-2),
(3,3),
(4,7),
(5,0),
(6,1),
(23,-5),
(45,4),
(78,-1),
(79,9),
(80,8),
(81,2),
(32111,6),
(58999,5)
}.
HERE INSERTION AT PREOCCUPIED INDEX IS ON SAME INDEX WITH SHIFTING ALL ELEMENTS ON and BEYOND THAT INDEX IN WORST-CASE OF (log m) time complexity.
2.> delete(int index):
It deletes data ,if present , at index . Worst-Case Time is O(log m).
3.> getAt(int index)::
It retrieves data ,if present, at index.Time in Worst-Case is O(log m).
4.> printAll():: It will print All elements(data) in increasing order of index.Time in ALL CASES is O(m),m= no. of elements present.
TIME COST::
Here Worst Case Time for each of Ist 3 operations is O(log m) at any time where m=no. of elements present at that time and for last one is O(m).
SPACE COST::
Space used is O(m), m= no. of elements present.
AS I HAVE THOROUGHLY SEARCHED ON INTERNET BUT NOT FOUND SOLUTIONS WITH SUCH OPTIMIZED TIME AND SPACE COST FOR ALL ABOVE STATED 4 OPERATIONS.
I WANT TO KNOW WETHER SUCH TIME AND MEMORY COST HAS BEEN ACHIEVED FOR ALL ABOVE 4 OPERATIONS BY ANYONE TILL NOW ,if not Can it be patented?
Also I cannot reveal anything about my algorithm more than this ....
I don't think that's possible.
If we use ordered binary search tree then also for the above test cases it will take time as O(log 58999!) and space to be O(58999).
It is because all the empty nodes would have to be created for reaching last index .
Even if we use array then also we have to create array till 58999 index and moreover worst case insertion will take O(58989) time complexity.
#Gaara :Its amazing that you have achieved O(log 15) time complexity in worst case but for getting a patent I truely don't know.As you will not reveal anything more about your algorithm...
#David Eisenstat: I also want to know how you will achieve it .
I have a Big O notation question. Say I have a Java program that does the following things:
Read an Array of Integers into a HashMap that keeps track of how many occurrences of the Integers exists in the array. [1,2,3,1] would be [1->2, 2->1, 3->1].
Then I grab the Keys from the HashMap and place them in an Array:
Set<Integer> keys = dictionary.keySet();
Integer[] keysToSort = new Integer[keys.size()];
keys.toArray(keysToSort);
Sort the keyArray using Arrays.sort.
Then iterate through the sorted keyArray grabbing the corresponding value from the HashMap, in order to display or format the results.
I think I know the following:
Step 1 is O(n)
Step 3 is O(n log n) if I'm to believe the Java API
Step 4 is O(n)
Step 2: When doing this type of calculation I should know how Java implements the Set class toArray method. I would assume that it iterates through the HashMap retrieving the Keys. If that's the case I'll assume its O(n).
If sequential operations dictate I add each part then the final calculation would be
O(n + n·log n + n+n) = O(3n+n·log n).
Skip the constants and you have O(n+n log n). Can this be reduced any further or am I just completely wrong?
I believe O(n + nlogn) can be further simplified to just O(nlogn). This is because the n becomes asymptotically insignificant compared to the nlogn because they are different orders of complexity. The nlogn is of a higher order than n. This can be verified on the wikipedia page by scrolling down to the Order of Common Functions section.
When using complex data structures like hash maps you do need to know how it retrieves the object, not all data structures have the same retrieval process or time to retrieve elements.
This might help you with the finding the Big O of complex data types in Java:
http://www.coderfriendly.com/wp-content/uploads/2009/05/java_collections_v2.pdf
Step 2 takes O(capacity of the map).
Step 1 and 4 can get bad if you have many keys with same hash code (i.e. O(number of those keys) for a single lookup or change, multiply with the number of those lookups/changes).
O(n + n·log n) = O(n·log n)
You are correct to worry a little about step 2. As far as I can tell the Java API does not specify running times for these operations.
As for O(n + n log n) Treebranch is right. You can reduce that to O(n log n) the reason being that for some base value n0 n log n > c*n forall c /= 0, n > n0 this is obviously the case, since no matter what number you chose for c you could use an n0 set to 2^c+1
First,
Step 1 is only O(n) if inserting integers into a HashMap is O(1). In Perl, the worse case for inserting into a hash is O(N) for N items (aka amortised O(1)), and that's if you discount the length of the key (which is acceptable here). HashMap could be less efficient depending on how it addresses certain issues.
Second,
O(N) is O(N log N), so O(N + N log N) is O(N log N).
One thing big O doesn't tell you is that how big the scaling factor is. It also assume you have an ideal machine. The reason this is imporant is that read from a file is likely to be far more expensive than everything else you do.
If you actually time this you will get something which is startup cost + read time. The startup cost is likely to be the largest for even one million records. The read time will be propertional to the number of bytes read (i.e. the length of the numbers can matter) If you have 100 million the read time is likely to be more important. If you have one billion records, alot will depend on the number of unique entries rather than the total number of entries. The number of unique entries is limited to ~2 billion.
BTW: To perform the counting more efficiently, try TIntIntHashMap which can minimise object creation making it several times faster.
Of course I am only talking about real machines which big O doesn't consider ;)
The point I am making is that you can do a big O calculation but it will not be informative as to how a real application will behave.
So I have a physical chart of time intervals (minute:second) which map to point values (for example: 9:59-10:10 = 59.7) and I need to write a program that tries to find out the point value for a given time (such as 10:02 would return 59.7).
I would also like to have the interval chart stored in a .properties file, so my other "calculators" are all consistent.
What would be the best way to program this?
One simple way would be to map the time to an integer - either number of seconds through the day (minutes * 60 + seconds) or just effectively "remove the colon" mapping 9:59 to 959, and 10:10 to 1010 (minutes * 100 + seconds).
Then each interval is just a pair of integers. If you have lots of intervals you may want to store them in a sorted list and perform a binary chop - if you don't have very many of them (or don't need to do this very often) then simply having a list of interval/value and walking through the list would be pretty simple.