With given N lists of M numbers in each list we have to find ONE element from each group such
every pair ai aj gives |ai-aj| as small as possible.
For example
we have 3 lists
{12,16,67,43}
{7,17,68,48}
{14,15,77,54}
And to minimize result we have to pick
number 16 from list 1
number 17 from list 2
number 15 from list 3
so
|16-17|=1
|16-15|=1
|17-15|=2
so our result is :2
How to solve it fastly? in N*M time ? or log something time
Chris
If you use a linear search, the complexity is O(N*M) for one match (i.e., for each element in set j, do a linear search for the most similar item from set i, and pick the smallest of those results.
If you sort each set first, you get to (at least) O(N log N)+O(M log M) for the sort, and O(M log N) for the searches (where N is the number of elements in set i, and M the number of elements in set j). If you walk through the two sets together you can probably reduce that to O(N + M) for the combined search.
Related
Maybe I'm missing some simple answer. Task - we have a set of size n*k, need to get a list of lists (size n) of sets (size k) with all possible combinations for splitting the set into parts of this size.
Sample:
Set:
(1,2,3,4,5,6)
Size: 3
Expected result:
[[(1,2,3),(4,5,6)],
[(1,2,4),(3,5,6)],
[(1,2,5),(3,4,6)],
[(1,2,6),(3,4,5)],
[(1,3,4),(2,5,6)],
[(1,3,5),(2,4,6)],
[(1,3,6),(2,4,5)],
[(1,4,5),(2,3,6)],
[(1,4,6),(2,3,5)],
[(1,5,6),(2,3,4)]]
The only thing that comes to mind is to find all subsets of size k and group them into lists taking only these that contain non-repeating elements.
The problem statement is ->
We want to scan N documents using two scanners. If S1 and S2 are the time taken by the scanner 1 and scanner 2 to scan a single document, find the minimum time required to scan all the N documents.
Example:-
S1 = 2, S2 = 4, N = 2
Output: 4
Explanation: Here we have two possibilities.
Either scan both documents in scanner 1 or
scan one document in each scanner.
In both the cases time required is 4.
I came up with a solution where we find all the combinations and insert them into the set. The minimum value will be the first element in the set.
The problem is the solution will have time complexity of O(n), but the accepted time complexity is O(logn).
I am thinking on the lines of binary search but can't come up with a solution.
What should be the approach?
If you could scan a fraction of a document, the fastest way would be to scan N*S2/(S1+S2) documents in scanner 1, and N*S1/(S1+S2) documents in scanner 2. And if these are not integers, you must round one of them up and one of them down, which gives you just two possibilities to check. This is O(1), not O(log n).
Well, I'm sharing the O(log n) approach. With binary search on ans / time, we could find the optimal time.
For binary search, we need upper bound & lower bound. Let's assume lower bound as 0. Now we need to find out the upper bound.
What will be the minimum time required if we scan all the documents in one scanner. It will be min (S1,S2) * N, right? Note: here we are not using other scanner which could scan documents while another one is busy. So min(S1,S2) * N will be our upper bound.
We've got our bounds,
Lower bound = 0
Upper bound = min(S1,S2) * N
Now do BS on time, take a mid & check how many documents can be scanned with scanner 1 scanner 2 within mid time. Whenever total scanned documents get >= N then mid could be ans
.
You can check BS from here - https://www.hackerearth.com/practice/algorithms/searching/binary-search/tutorial/
Given a list of N intervals [a,b] cost c, find the minimum sum of 2 non-overlapping intervals. I have an algorithm in O(n^2) ( pastebin.com/kveAZTwv ) but i don t find the O(N log N).
The first value is the number of intervals. the other lines are : a,b,c where a is the beginning of the interval, b the end and c the cost.
example :
input :
3
0 10 1
1 2 2
9 12 2
output :
4
Here is the basic idea for an algorithm in O(n log n), I am sure it can be done more efficiently:
1) sort all intervals accoring to their endpoint, each endpoint is a potential splitpoint where intervals to the right and intervals to the left do not overlap
2) now scan through the sorted intervals and for each spiltpoint remember the minimum cost interval to the left.
3) sort all elements according to their startpoint
4) for each splitpoint memorize additionally the minimum cost interval to the right(having its startpoint after the splitpoint). This is also possible with a single scan from back to front of the sorted elements
5) for each splitpoint add the two corresponding cost and look for the minimum.
Sorry this is a bit informal since i am on mobile.
I recently came across an interesting problem:
Given two lists of intervals, find the total number of overlapping intervals from the two lists.
Example
L1: ([1,2][2,3][4,5][6,7])
L2: ([1,5][2,3][4,7][5,7])
[1,5] overlaps [1,2] [2,3] [4,5]
[2,3] overlaps [1,2] [2,3]
[4,7] overlaps [4,5] [6,7]
[5,7] overlaps [4,5] [6,7]
total = 3+2+2+2 = 9
Obviously the brute force approach works, but it's too slow (I need something better than O(n^2)).
I also fond a similar problem here. But it's not exactly the same...
Any help is appreciated
Make two sorted lists with pairs (value; +1 or -1 for start and end of interval).
Two counters - Count1 and Count2 which show number of active intervals in the first and the second lists.
Walk through both lists in merge manner.
When you get pair from the first list with +1 - increment Count1
When you get pair from the first list with -1 - decrement Count1 and add Count2 to the result
The same for pairs from the second list
Pseudocode for the last stage
CntA = 0
CntB = 0
Res = 0
ia = 0
ib = 0
while (ia < A.Length) and (ib < B.Length)
if Compare(A[ia], B[ib]) <= 0
CntA = CntA + A[ia].Flag
if (A[ia].Flag < 0)
Res = Res + CntB
ia++
else
CntB = CntB + B[ib].Flag
if B[ib].Flag < 0
Res = Res + CntA
ib++
Subtle moment - comparison if Compare(A[ia], B[ib]) <= 0
We should here take into account also flags - to correctly treat situations when endpoints only touch like [1..2][2..3] (you consider this situation as intersection). So both sorting and merge comparator should take synthetic value like this: 3 * A[ia].Value - A[ia].Flag. With such comparing start of interval is treated before end of interval with the same coordinate.
P.S. Made quick test in Delphi. Works for given data set and pair of others.
Delphi code (ideone FPC doesn't compile it due to generics)
Try to look for sweep line algorithms, it will give you the fastest solution.
You can check short description at TopCoder site or watch video from Robert Sedgwick. These describe a bit more hard problem but should give you an approach how to solve yours.
Actually the main idea is to walk over sorted list of begins and ends of your segments each time updating the lists of segments in the special intersecting list.
For this task you will have two intersections lists for each original list respectively. At the start both intersection lists are empty. On coming over begin of the segment you add it to the appropriate intersection list and it obviously intersects all the segments in the other intersection list. When coming to an end of the segment just remove it from the intersection list.
This algorithm will give you O(n log(n)) speed and in worst case O(n) memory.
You may be able to use std::set_intersection in a loop over the second array to match it with each item in the first array. But I am not sure if the performance will match your requirements.
I recently stumbled upon the Interval Tree ADT when tackling a similar question - I suspect it'll be useful for you, whether you implement it or not.
It is basically a ternary tree, and I built it with nodes containing:
Left sub-tree containing intervals less than current node
Right sub-tree containing intervals more than current node
List of overlapping intervals
Interval value encompassing all overlapping intervals
After building the tree in O(n*log(n)), a query function - to check overlapping intervals - should be O(log(n) + m) where m is the number of overlapping intervals reported.
Note that on creation, sorting by end value in the interval and splitting the list should help keep things balanced.
This is my first post. I have been teaching myself Prolog for a university project and I am tasked with generating a program that simulates the lotto and compares the random numbers (6 in this case) with the numbers that the user has. If they all match then you are deemed the winner if not, then it returns 'hard luck'.
All I have been able to do so far is generate one random number in the range of 1-50. I don't know how to do much else after that.
:- use_module(library(random)).
?- random(1,50,Out).
I understand I have to add the random number to a list, but I'm not sure how to implement it. And to then have another list of numbers (user_numbers) in the database or fact-base. Then use SWI-Prolog to check if they're equal.
It is a really tough program for me to try and do in Prolog, especially seeing as I am teaching it to myself. If anybody could give me some pointers on how to approach it I would be very grateful.
pick_number(N) :- random(1, 50, N).
We need to pick a list of 6 numbers
lotto_numbers(Ns) :-
length(Ns, 6), % The length of our list is 6, now we have a list of 6 free variables.
select_numbers(Ns). % We need to select our numbers, binding them to our free variables.
select_numbers([]). % An empty list needs no picking
select_numbers([N|Ns]) :-
pick_number(N), % We pick the first number (bind the free variable to a random number)
select_numbers(Ns). % Then we pick the rest of the numbers.
We need to check if the ticket holder has winning numbers. Does it matter what order the numbers are in? If so, then we can check if the two lists unify: LottoNumbers = LottoTicketNumbers. If we don't care about order, then we need a slightly more complex solution:
numbers_match([], []). % if both lists are empty, then they must have all matched.
numbers_match([N|Ns], Ms) :-
select(N, Ms, NewMs), % remove N from Ms (if N matches an element in Ms), leaving NewMs
numbers_match(Ns, NewMs). % remove the rest of Ns from NewMs.
If both lists don't empty at the same time, then they didn't all match up.
Supposing we have some loto ticket in the database,
lotto_ticket(Ns) :- lotto_numbers(Ns).
With all the above definitions in our program, we can generate a lotto ticket, and generate some lotto numbers (actually the same process, but named differently for illustrative purposes), and see if they have all and only the same numbers:
?- lotto_ticket(T), lotto_numbers(L), numbers_match(T, L).
false.
Ah. No surprise that we lost...
That's all fine and good, but we can save a lot of steps by using a higher-order
predicate and some common library predicates:
alt_lotto_numbers(Ns) :-
length(Ns, 6),
maplist(random(1,50), Ns). % `maplist/2` is just a way of calling a predicate on every member of a list.
alt_numbers_match(Ns, Ms) :-
same_length(Ns, Ms),
subset(Ns, Ms).