This was one of the interview questions, and I was wondering if I did it right, and want to know if my solution is the most efficient one. O(n log n)
Given a bag of nuts and a bag of bolts, each having a different size within a bag but exactly one match in the other bag, give a fast algorithm to find all matches.
Since there is always a match between 2 bags, I said there will be same number of nuts and same number of bolts. Let the number be n.
I would first sort components in each bag each component's weight, and it is possible to do that because they all have different size within a bag. Using merge sort, it will have O(n log n) time complexity.
Next, it would be just easy process of matching each component in 2 bags from lightest to heaviest.
I want to know if this is the right solution, and also if there is any other interesting way to solve this problem.
There is a solution to this problem in O(nlgn) that works even if you can't compare two nuts or two bolts directly. That is without using a scale, or if the differences in weight are too small to observe.
It works by applying a small variation of randomised quicksort as follows:
Pick a random bolt b.
Compare bolt b with all the nuts, partitioning the nuts into those of size less than b and greater than b.
Now we must also partition the bolts into two halves as well and we can't compare bolt to bolt. But now we know what is the matching nut n to b. So we compare n to all the bolts, partitioning them into those of size less than n and greater than n.
The rest of the problem follows directly from randomised quicksort by applying the same steps to the lessThan and greaterThan partitions of nuts and bolts.
The standard solution is to put the bag of nuts into a hash (or HashMap, or dictionary, or whatever your language of choice calls it), and then walk the other bag using hash lookups to find the matches.
Average performance of this algorithm with be O(n), with better constants than your sort/binary search variation.
Assuming you have no exact measurements of size, and you can only see if something is too large, too small, or exactly right by trying, this needs the extra step of how you are sorting -- which is probably by taking a random nut, comparing with each bolt and sorting into smaller, bigger piles, and then taking a bolt and sorting the nuts with that (a modified quicksort... keep in mind you'll have to deal with the problem of how to choose a nut or bolt to compare to when the piles get smaller as to make sure you're using a reasonable pivot).
Since the problem doesn't tell you that the nuts and bolts are labeled with sizes, or that you have some way of measuring the nuts and bolts' sizes, it would be difficult to use a hash table to solve this, or even a regular comparison sort.
That looks like the best solution to me, complexity-wise.
An alternative would be to sort only one of the bags, then use binary search to try matching bolt to nut (removing the pair from the sorted list). Still O(n log n), but the second part is optimized slightly, since we're removing each pair after the match, the worst case would be: log n + log n-1 + log n-2 ... 1 = (n log n)/2
Take two bags extra. Say left and right bags. Now take each nut and compare with each bolt. If the bolt is small place it in left bag, large then in right bag, fits then place it away. Now take the next nut, compare it with previously taken nut, if its smaller, then we need to check only those nuts in left bag, else to the ones in right bag. Continue in this fashion.
Related
This problem is 4-11 of Skiena. The solution to finding majority elements - repeated more than half times is majority algorithm. Can we use this to find all numbers repeated n/4 times?
Misra and Gries describe a couple approaches. I don't entirely understand their paper, but a key idea is to use a bag.
Boyer and Moore's original majority algorithm paper has a lot of incomprehensible proofs and discussion of formal verification of FORTRAN code, but it has a very good start of an explanation of how the majority algorithm works. The key concept starts with the idea that if the majority of the elements are A and you remove, one at a time, a copy of A and a copy of something else, then in the end you will have only copies of A. Next, it should be clear that removing two different items, neither of which is A, can only increase the majority that A holds. Therefore it's safe to remove any pair of items, as long as they're different. This idea can then be made concrete. Take the first item out of the list and stick it in a box. Take the next item out and stick it in the box. If they're the same, let them both sit there. If the new one is different, throw it away, along with an item from the box. Repeat until all items are either in the box or in the trash. Since the box is only allowed to have one kind of item at a time, it can be represented very efficiently as a pair (item type, count).
The generalization to find all items that may occur more than n/k times is simple, but explaining why it works is a little harder. The basic idea is that we can find and destroy groups of k distinct elements without changing anything. Why? If w > n/k then w-1 > (n-k)/k. That is, if we take away one of the popular elements, and we also take away k-1 other elements, then the popular element remains popular!
Implementation: instead of only allowing one kind of item in the box, allow k-1 of them. Whenever you see a group of k different items show up (that is, there are k-1 types in the box, and the one arriving doesn't match any of them), you throw one of each type in the trash, including the one that just arrived. What data structure should we use for this "box"? Well, a bag, of course! As Misra and Gries explain, if the elements can be ordered, a tree-based bag with O(log k) basic operations will give the whole algorithm a complexity of O(n log k). One point to note is that the operation of removing one of each element is a bit expensive (O(k) for a typical implementation), but that cost is amortized over the arrivals of those elements, so it's no big deal. Of course, if your elements are hashable rather than orderable, you can use a hash-based bag instead, which under certain common assumptions will give even better asymptotic performance (but it's not guaranteed). If your elements are drawn from a small finite set, you can guarantee that. If they can only be compared for equality, then your bag gets much more expensive and I'm pretty sure you end up with something like O(nk) instead.
Find the majority element that appears n/2 times by Moore-Voting Algorithm
See method 3 of the given link for Moore's Voting Algo (http://www.geeksforgeeks.org/majority-element/).
Time:O(n)
Now after finding majority element, scan the array again and remove the majority element or make it -1.
Time:O(n)
Now apply Moore Voting Algorithm on the remaining elements of array (but ignore -1 now as it has already been included earlier). The new majority element appears n/4 times.
Time:O(n)
Total Time:O(n)
Extra Space:O(1)
You can do it for element appearing more than n/8,n/16,.... times
EDIT:
There may exist a case when there is no majority element in the array:
For e.g. if the input arrays is {3, 1, 2, 2, 1, 2, 3, 3} then the output should be [2, 3].
Given an array of of size n and a number k, find all elements that appear more than n/k times
See this link for the answer:
https://stackoverflow.com/a/24642388/3714537
References:
http://www.cs.utexas.edu/~moore/best-ideas/mjrty/
See this paper for a solution that uses constant memory and runs in linear time, which will find 3 candidates for elements that occur more than n/4 times. Note that if you assume that your data is given as a stream that you can only go through once, this is the best you can do -- you have to go through the stream one more time to test each of the 3 candidates to see if it occurs more than n/4 times in the stream. However, if you assume a priori that there are 3 elements that occur more than n/4 times then you only need to go through the stream once so you get a linear time online algorithm (only goes through the stream once) that only requires constant storage.
As you didnt mention space complexity , one possible solution is using hashtable for the elements which maps to count then you can just increment count if the element is found.
I have encountered an algorithmic problem but am not able to figure out anything better than brute force or reduce it to a better know problem. Any hints?
There are N bags of variable sizes and N types of items. Each type of items belongs to one bag. There are lots of items of each type and each item may be of a different size. Initially, these items are distributed across all the bags randomly. We have to place the items in their respective bags. However, we can only operate with a pair of bags at one time by exchanging items (as much as possible) and proceeding to the next pair. The aim is to reduce the total number of pairs. Edit: The aim is to find a sequence of transfers that minimizes the total number of bag pairs involved
Clarification:
The bags are not arbitrarily large (You can assume the bag and item sizes to be integers between 0 to 1000 if it helps). You'll frequently encounter scenarios where the all the items between 2 bags cannot be swapped due to the limited capacity of one of the bags. This is where the algorithm needs to make an optimisation. Perhaps, if another pair of bags were swapped first, the current swap can be done in one go. To illustrate this, let's consider Bags A, B and C and their items 1, 2, 3 respectively. The number in the brackets is the size.
A(10) : 3(8)
B(10): 1(2), 1(3)
C(10): 1(4)
The swap orders can be AB, AC, AB or AC, AB. The latter is optimal as the number of swaps is lesser.
Since I cannot come to an idea for an algorithm that will always find an optimal answer, and approximation of the fitness of the solution (amount of swaps) is also fine, I suggest a stochastic local search algorithm with pruning.
Given a random starting configuration, this algorithm considers all possible swaps, and makes a weighed decision based on chance: the better a swap is, the more likely it is chosen.
The value of a swap would be the sum of the value of the transaction of an item, which is zero if the item does not end up in it's belonging bag, and is positive if it does end up there. The value increases as the item's size increases (the idea behind this is that a larger block is hard to move many times in comparison to smaller blocks). This fitness function can be replaced by any other fitness function, it's efficiency is unknown until empirically shown.
Since any configuration can be the consequence of many preceding swaps, we keep track of which configurations we have seen before, along with a fitness (based on how many items are in their correct bag - this fitness is not related to the value of a swap) and the list of preceded swaps. If the fitness function for a configuration is the sum of the items that are in their correct bags, then the amount of items in the problem is the highest fitness (and therefor marks a configuration to be a solution).
A swap is not possible if:
Either of the affected bags is holding more than it's capacity after the potential swap.
The new swap brings you back to the last configuration you were in before the last swap you did (i.e. reversed swap).
When we identify potential swaps, we look into our list of previously seen configurations (use a hash function for O(1) lookup). Then we either set its preceded swaps to our preceded swaps (if our list is shorter than it's), or we set our preceded swaps to its list (if it's list is shorter than ours). We can do this because it does not matter which swaps we did, as long as the amount of swaps is as small as possible.
If there are no more possible swaps left in a configuration, it means you're stuck. Local search tells you 'reset' which you can do in may ways, for instance:
Reset to a previously seen state (maybe the best one you've seen so far?)
Reset to a new valid random solution
Note
Since the algorithm only allows you to do valid swaps, all constraints will be met for each configuration.
The algorithm does not guarantee to 'stop' out of the box, you can implement a maximum number of iterations (swaps)
The algorithm does not guarantee to find a correct solution, as it does it's best to find a better configuration each iteration. However, since a perfect solution (set of swaps) should look closely to an almost perfect solution, a human might be able to finish what the local search algorithm was not after it results in a invalid configuration (where not every item is in its correct bag).
The used fitness functions and strategies are very likely not the most efficient out there. You could look around to find better ones. A more efficient fitness function / strategy should result in a good solution faster (less iterations).
Came across this interview question.
Write an algorithm to find the mean(average) of a large list. This
list could contain trillions or quadrillions of number. Each number is
manageable in hundreds, thousands or millions.
Googling it gave me all Median of Medians solutions. How should I approach this problem?
Is divide and conquer enough to deal with trillions of number?
How to deal with the list of the such a large size?
If the size of the list is computable, it's really just a matter of how much memory you have available, how long it's supposed to take and how simple the algorithm is supposed to be.
Basically, you can just add everything up and divide by the size.
If you don't have enough memory, dividing first might work (Note that you will probably lose some precision that way).
Another approach would be to recursively split the list into 2 halves and calculating the mean of the sublists' means. Your recursion termination condition is a list size of 1, in which case the mean is simply the only element of the list. If you encounter a list of odd size, make either the first or second sublist longer, this is pretty much arbitrary and doesn't even have to be consistent.
If, however, you list is so giant that its size can't be computed, there's no way to split it into 2 sublists. In that case, the recursive approach works pretty much the other way around. Instead of splitting into 2 lists with n/2 elements, you split into n/2 lists with 2 elements (or rather, calculate their mean immediately). So basically, you calculate the mean of elements 1 and 2, that becomes you new element 1. the mean of 3 and 4 is your new second element, and so on. Then apply the same algorithm to the new list until only 1 element remains. If you encounter a list of odd size, either add an element at the end or ignore the last one. If you add one, you should try to get as close as possible to your expected mean.
While this won't calculate the mean mathematically exactly, for lists of that size, it will be sufficiently close. This is pretty much a mean of means approach. You could also go the median of medians route, in which case you select the median of sublists recursively. The same principles apply, but you will generally want to get an odd number.
You could even combine the approaches and calculate the mean if your list is of even size and the median if it's of odd size. Doing this over many recursion steps will generate a pretty accurate result.
First of all, this is an interview question. The problem as stated would not arise in practice. Also, the question as stated here is imprecise. That is probably deliberate. (They want to see how you deal with solving an imprecisely specified problem.)
Write an algorithm to find the mean(average) of a large list.
The word "find" is rubbery. It could mean calculate (to some precision) or it could mean estimate.
The phrase "large list" is rubbery. If could mean a list or array data structure in memory, or the "list" could be the result of a database query, the contents of a file or files.
There is no mention of the hardware constraints on the system where this will be implemented.
So the first thing >>I<< would do would be to try to narrow the scope by asking some questions of the interviewer.
But assuming that you can't, then a complete answer would need to cover the following points:
The dataset probably won't fit in memory at the same time. (But if it does, then that is good.)
Calculating the average of N numbers is O(N) if you do it serially. For N this size, it could be an intractable problem.
An alternative is to split into sublists of equals size and calculate the averages, and the average of the averages. In theory, this gives you O(N/P) where P is the number of partitions. The parallelism could be implemented with multiple threads, with multiple processes on the same machine, or distributed.
In practice, the limiting factors are going to be computational, memory and/or I/O bandwidth. A parallel solution will be effective if you can address these limits. For example, you need to balance the problem of each "worker" having uncontended access to its "sublist" versus the problem of making copies of the data so that that can happen.
If the list is represented in a way that allows sampling, then you can estimate the average without looking at the entire dataset. In fact, this could be O(C) depending on how you sample. But there is a risk that your sample will be unrepresentative, and the average will be too inaccurate.
In all cases doing calculations, you need to guard against (integer) overflow and (floating point) rounding errors. Especially while calculating the sums.
It would be worthwhile discussing how you would solve this with a "big data" platform (e.g. Hadoop) and the limitations of that approach (e.g. time taken to load up the data ...)
I'm writing a program for a competition and I need to be faster than all the other competitors. For this I need a little algorithm help; ideally I'd be using the fastest algorithm.
For this problem I am given 2 things. The first is a list of tuples, each of which contains exactly two elements (strings), each of which represents an item. The second is an integer, which indicates how many unique items there are in total. For example:
# of items = 3
[("ball","chair"),("ball","box"),("box","chair"),("chair","box")]
The same tuples can be repeated/ they are not necessarily unique.) My program is supposed to figure out the maximum number of tuples that can "agree" when the items are sorted into two groups. This means that if all the items are broken into two ideal groups, group 1 and group 2, what are the maximum number of tuples that can have their first item in group 1 and their second item in group 2.
For example, the answer to my earlier example would be 2, with "ball" in group 1 and "chair" and "box" in group 2, satisfying the first two tuples. I do not necessarily need know what items go in which group, I just need to know what the maximum number of satisfied tuples could be.
At the moment I'm trying a recursive approach, but its running on (n^2), far too inefficient in my opinion. Does anyone have a method that could produce a faster algorithm?
Thanks!!!!!!!!!!
Speed up approaches for your task:
1. Use integers
Convert the strings to integers (store the strings in an array and use the position for the tupples.
String[] words = {"ball", "chair", "box"};
In tuppls ball now has number 0 (pos 0 in array) , chair 1, box 2.
comparing ints is faster than Strings.
2. Avoid recursion
Recursion is slow, due the recursion overhead.
For example look at binarys search algorithm in a recursive implementatiion, then look how java implements binSearch() (with a while loop and iteration)
Recursion is helpfull if problems are so complex that a non recursive implementation is to complex for a human brain.
An iterataion is faster, but not in the case when you mimick recursive calls by implementing your own stack.
However you can start implementing using a recursiove algorithm, once it works and it is a suited algo, then try to convert to a non recursive implementation
3. if possible avoid objects
if you want the fastest, the now it becomes ugly!
A tuppel array can either be stored in as array of class Point(x,y) or probably faster,
as array of int:
Example:
(1,2), (2,3), (3,4) can be stored as array: (1,2,2,3,3,4)
This needs much less memory because an object needs at least 12 bytes (in java).
Less memory becomes faster, when the array are really big, then your structure will hopefully fits in the processor cache, while the objects array does not.
4. Programming language
In C it will be faster than in Java.
Maximum cut is a special case of your problem, so I doubt you have a quadratic algorithm for it. (Maximum cut is NP-complete and it corresponds to the case where every tuple (A,B) also appears in reverse as (B,A) the same number of times.)
The best strategy for you to try here is "branch and bound." It's a variant of the straightforward recursive search you've probably already coded up. You keep track of the value of the best solution you've found so far. In each recursive call, you check whether it's even possible to beat the best known solution with the choices you've fixed so far.
One thing that may help (or may hurt) is to "probe": for each as-yet-unfixed item, see if putting that item on one of the two sides leads only to suboptimal solutions; if so, you know that item needs to be on the other side.
Another useful trick is to recurse on items that appear frequently both as the first element and as the second element of your tuples.
You should pay particular attention to the "bound" step --- finding an upper bound on the best possible solution given the choices you've fixed.
I want to sort items where the comparison is performed by humans:
Pictures
Priority of work items
...
For these tasks the number of comparisons is the limiting factor for performance.
What is the minimum number of comparisons needed (I assume > N for N items)?
Which algorithm guarantees this minimum number?
To answer this, we need to make a lot of assumptions.
Let's assume we are sorting pictures by cuteness. The goal is to get the maximum usable information from the human in the least amount of time. This interaction will dominate all other computation, so it's the only one that counts.
As someone else mentioned, humans can deal well with ordering several items in one interaction. Let's say we can get eight items in relative order per round.
Each round introduces seven edges into a directed graph where the nodes are the pictures. If node A is reachable from node B, then node A is cuter than node B. Keep this graph in mind.
Now, let me tell you about a problem the Navy and the Air Force solve differently. They both want to get a group of people in height order and quickly. The Navy tells people to get in line, then if you're shorter than the guy in front of you, switch places, and repeat until done. In the worst case, it's N*N comparison.
The Air Force tells people to stand in a square grid. They shuffle front-to-back on sqrt(N) people, which means worst case sqrt(N)*sqrt(N) == N comparisons. However, the people are only sorted along one dimension. So therefore, the people face left, then do the same shuffle again. Now we're up to 2*N comparisons, and the sort is still imperfect but it's good enough for government work. There's a short corner, a tall corner opposite, and a clear diagonal height gradient.
You can see how the Air Force method gets results in less time if you don't care about perfection. You can also see how to get the perfection effectively. You already know that the very shortest and very longest men are in two corners. The second-shortest might be behind or beside the shortest, the third shortest might be behind or beside him. In general, someone's height rank is also his maximum possible Manhattan distance from the short corner.
Looking back at the graph analogy, the eight nodes to present each round are eight of those with the currently most common length of longest inbound path. The length of the longest inbound path also represents the node's minimum possible sorted rank.
You'll use a lot of CPU following this plan, but you will make the best possible use of your human resources.
From an assignment I once did on this very subject ...
The comparison counts are for various sorting algorithms operating on data in a random order
Size QkSort HpSort MrgSort ModQk InsrtSort
2500 31388 48792 25105 27646 1554230
5000 67818 107632 55216 65706 6082243
10000 153838 235641 120394 141623 25430257
20000 320535 510824 260995 300319 100361684
40000 759202 1101835 561676 685937
80000 1561245 2363171 1203335 1438017
160000 3295500 5045861 2567554 3047186
These comparison counts are for various sorting algorithms operating on data that is started 'nearly sorted'. Amongst other things it shows a the pathological case of quicksort.
Size QkSort HpSort MrgSort ModQk InsrtSort
2500 72029 46428 16001 70618 76050
5000 181370 102934 34503 190391 3016042
10000 383228 226223 74006 303128 12793735
20000 940771 491648 158015 744557 50456526
40000 2208720 1065689 336031 1634659
80000 4669465 2289350 712062 3820384
160000 11748287 4878598 1504127 10173850
From this we can see that merge sort is the best by number of comparisons.
I can't remember what the modifications to the quick sort algorithm were, but I believe it was something that used insertion sorts once the individual chunks got down to a certain size. This sort of thing is commonly done to optimise quicksort.
You might also want to look up Tadao Takaoka's 'Minimal Merge Sort', which is a more efficient version of the merge sort.
Pigeon hole sorting is order N and works well with humans if the data can be pigeon holed. A good example would be counting votes in an election.
You should consider that humans might make non-transitive comparisons, e.g. they favor A over B, B over C but also C over A. So when choosing your sort algorithm, make sure it doesn't completely break when that happens.
People are really good at ordering 5-10 things from best to worst and come up with more consistent results when doing so. I think trying to apply a classical sorting algo might not work here because of the typically human multi-compare approach.
I'd argue that you should have a round robin type approach and try to bucket things into their most consistent groups each time. Each iteration would only make the result more certain.
It'd be interesting to write too :)
If comparisons are expensive relative to book-keeping costs, you might try the following algorithm which I call "tournament sort". First, some definitions:
Every node has a numeric "score" property (which must be able to hold values from 1 to the number of nodes), and a "last-beat" and "fellow-loser" properties, which must be able to hold node references.
A node is "better" than another node if it should be output before the other.
An element is considered "eligible" if there are no elements known to be better than it which have been output, and "ineligible" if any element which has not been output is known to be better than it.
The "score" of a node is the number of nodes it's known to be better than, plus one.
To run the algorithm, initially assign every node a score of 1. Repeatedly compare the two lowest-scoring eligible nodes; after each comparison, mark the loser "ineligible", and add the loser's score to the winner's (the loser's score is unaltered). Set the loser's "fellow loser" property to the winner's "last-beat", and the winner's "last-beat" property to the loser. Iterate this until only one eligible node remains. Output that node, and make eligible all nodes the winner beat (using the winner's "last-beat" and the chain of "fellow-loser" properties). Then continue the algorithm on the remaining nodes.
The number of comparisons with 1,000,000 items was slightly lower than that of a stock library implementation of Quicksort; I'm not sure how the algorithm would compare against a more modern version of QuickSort. Bookkeeping costs are significant, but if comparisons are sufficiently expensive the savings could possibly be worth it. One interesting feature of this algorithm is that it will only perform comparisons relevant to determining the next node to be output; I know of no other algorithm with that feature.
I don't think you're likely to get a better answer than the Wikipedia page on sorting.
Summary:
For arbitrary comparisons (where you can't use something like radix sorting) the best you can achieve is O(n log n)
Various algorithms achieve this - see the "comparison of algorithms" section.
The commonly used QuickSort is O(n log n) in a typical case, but O(n^2) in the worst case; there are often ways to avoid this, but if you're really worried about the cost of comparisons, I'd go with something like MergeSort or a HeapSort. It partly depends on your existing data structures.
If humans are doing the comparisons, are they also doing the sorting? Do you have a fixed data structure you need to use, or could you effectively create a copy using a balanced binary tree insertion sort? What are the storage requirements?
Here is a comparison of algorithms. The two better candidates are Quick Sort and Merge Sort. Quick Sort is in general better, but has a worse worst case performance.
Merge sort is definately the way to go here as you can use a Map/Reduce type algorithm to have several humans doing the comparisons in parallel.
Quicksort is essentially a single threaded sort algorithm.
You could also tweak the merge sort algorithm so that instead of comparing two objects you present your human with a list of say five items and ask him or her to rank them.
Another possibility would be to use a ranking system as used by the famous "Hot or Not" web site. This requires many many more comparisons, but, the comparisons can happen in any sequence and in parallel, this would work faster than a classic sort provided you have enough huminoids at your disposal.
The questions raises more questions really.
Are we talking a single human performing the comparisons? It's a very different challenge if you are talking a group of humans trying to arrange objects in order.
What about the questions of trust and error? Not everyone can be trusted or to get everything right - certain sorts would go catastrophically wrong if at any given point you provided the wrong answer to a single comparison.
What about subjectivity? "Rank these pictures in order of cuteness". Once you get to this point, it could get really complex. As someone else mentions, something like "hot or not" is the simplest conceptually, but isn't very efficient. At it's most complex, I'd say that google is a way of sorting objects into an order, where the search engine is inferring the comparisons made by humans.
The best one would be the merge sort
The minimum run time is n*log(n) [Base 2]
The way it is implemented is
If the list is of length 0 or 1, then it is already sorted.
Otherwise:
Divide the unsorted list into two sublists of about half the size.
Sort each sublist recursively by re-applying merge sort.
Merge the two sublists back into one sorted list.