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I was set a homework challenge as part of an application process (I was rejected, by the way; I wouldn't be writing this otherwise) in which I was to implement the following functions:
// Store a collection of integers
class IntegerCollection {
public:
// Insert one entry with value x
void Insert(int x);
// Erase one entry with value x, if one exists
void Erase(int x);
// Erase all entries, x, from <= x < to
void Erase(int from, int to);
// Return the count of all entries, x, from <= x < to
size_t Count(int from, int to) const;
The functions were then put through a bunch of tests, most of which were trivial. The final test was the real challenge as it performed 500,000 single insertions, 500,000 calls to count and 500,000 single deletions.
The member variables of IntegerCollection were not specified and so I had to choose how to store the integers. Naturally, an STL container seemed like a good idea and keeping it sorted seemed an easy way to keep things efficient.
Here is my code for the four functions using a vector:
// Previous bit of code shown goes here
private:
std::vector<int> integerCollection;
};
void IntegerCollection::Insert(int x) {
/* using lower_bound to find the right place for x to be inserted
keeps the vector sorted and makes life much easier */
auto it = std::lower_bound(integerCollection.begin(), integerCollection.end(), x);
integerCollection.insert(it, x);
}
void IntegerCollection::Erase(int x) {
// find the location of the first element containing x and delete if it exists
auto it = std::find(integerCollection.begin(), integerCollection.end(), x);
if (it != integerCollection.end()) {
integerCollection.erase(it);
}
}
void IntegerCollection::Erase(int from, int to) {
if (integerCollection.empty()) return;
// lower_bound points to the first element of integerCollection >= from/to
auto fromBound = std::lower_bound(integerCollection.begin(), integerCollection.end(), from);
auto toBound = std::lower_bound(integerCollection.begin(), integerCollection.end(), to);
/* std::vector::erase deletes entries between the two pointers
fromBound (included) and toBound (not indcluded) */
integerCollection.erase(fromBound, toBound);
}
size_t IntegerCollection::Count(int from, int to) const {
if (integerCollection.empty()) return 0;
int count = 0;
// lower_bound points to the first element of integerCollection >= from/to
auto fromBound = std::lower_bound(integerCollection.begin(), integerCollection.end(), from);
auto toBound = std::lower_bound(integerCollection.begin(), integerCollection.end(), to);
// increment pointer until fromBound == toBound (we don't count elements of value = to)
while (fromBound != toBound) {
++count; ++fromBound;
}
return count;
}
The company got back to me saying that they wouldn't be moving forward because my choice of container meant the runtime complexity was too high. I also tried using list and deque and compared the runtime. As I expected, I found that list was dreadful and that vector took the edge over deque. So as far as I was concerned I had made the best of a bad situation, but apparently not!
I would like to know what the correct container to use in this situation is? deque only makes sense if I can guarantee insertion or deletion to the ends of the container and list hogs memory. Is there something else that I'm completely overlooking?
We cannot know what would make the company happy. If they reject std::vector without concise reasoning I wouldn't want to work for them anyway. Moreover, we dont really know the precise requirements. Were you asked to provide one reasonably well performing implementation? Did they expect you to squeeze out the last percent of the provided benchmark by profiling a bunch of different implementations?
The latter is probably too much for a homework challenge as part of an application process. If it is the first you can either
roll your own. It is unlikely that the interface you were given can be implemented more efficiently than one of the std containers does... unless your requirements are so specific that you can write something that performs well under that specific benchmark.
std::vector for data locality. See eg here for Bjarne himself advocating std::vector rather than linked lists.
std::set for ease of implementation. It seems like you want the container sorted and the interface you have to implement fits that of std::set quite well.
Let's compare only isertion and erasure assuming the container needs to stay sorted:
operation std::set std::vector
insert log(N) N
erase log(N) N
Note that the log(N) for the binary_search to find the position to insert/erase in the vector can be neglected compared to the N.
Now you have to consider that the asymptotic complexity listed above completely neglects the non-linearity of memory access. In reality data can be far away in memory (std::set) leading to many cache misses or it can be local as with std::vector. The log(N) only wins for huge N. To get an idea of the difference 500000/log(500000) is roughly 26410 while 1000/log(1000) is only ~100.
I would expect std::vector to outperform std::set for considerably small container sizes, but at some point the log(N) wins over cache. The exact location of this turning point depends on many factors and can only reliably determined by profiling and measuring.
Nobody knows which container is MOST efficient for multiple insertions / deletions. That is like asking what is the most fuel-efficient design for a car engine possible. People are always innovating on the car engines. They make more efficient ones all the time. However, I would recommend a splay tree. The time required for a insertion or deletion is a splay tree is not constant. Some insertions take a long time and some take only a very a short time. However, the average time per insertion/deletion is always guaranteed to be be O(log n), where n is the number of items being stored in the splay tree. logarithmic time is extremely efficient. It should be good enough for your purposes.
The first thing that comes to mind is to hash the integer value so single look ups can be done in constant time.
The integer value can be hashed to compute an index in to an array of bools or bits, used to tell if the integer value is in the container or not.
Counting and and deleting large ranges could be sped up from there, by using multiple hash tables for specific integer ranges.
If you had 0x10000 hash tables, that each stored ints from 0 to 0xFFFF and were using 32 bit integers you could then mask and shift the upper half of the int value and use that as an index to find the correct hash table to insert / delete values from.
IntHashTable containers[0x10000];
u_int32 hashIndex = (u_int32)value / 0x10000;
u_int32int valueInTable = (u_int32)value - (hashIndex * 0x10000);
containers[hashIndex].insert(valueInTable);
Count for example could be implemented as so, if each hash table kept count of the number of elements it contained:
indexStart = startRange / 0x10000;
indexEnd = endRange / 0x10000;
int countTotal = 0;
for (int i = indexStart; i<=indexEnd; ++i) {
countTotal += containers[i].count();
}
Not sure if using sorting really is a requirement for removing the range. It might be based on position. Anyway, here is a link with some hints which STL container to use.
In which scenario do I use a particular STL container?
Just FYI.
Vector maybe a good choice, but it does a lot of re allocation, as you know. I prefer deque instead, as it doesn't require big chunk of memory to allocate all items. For such requirement as you had, list probably fit better.
Basic solution for this problem might be std::map<int, int>
where key is the integer you are storing and value is the number of occurences.
Problem with this is that you can not quickly remove/count ranges. In other words complexity is linear.
For quick count you would need to implement your own complete binary tree where you can know the number of nodes between 2 nodes(upper and lower bound node) because you know the size of tree, and you know how many left and right turns you took to upper and lower bound nodes. Note that we are talking about complete binary tree, in general binary tree you can not make this calculation fast.
For quick range remove I do not know how to make it faster than linear.
I'm struggling with a part of my college assignment. I have two subsets of std::set containers containing pointers to quite complex objects, but ordered by different criteria (which is why I can't use std::set_intersection()). I need to find elements that are contained in both subsets as fast as possible. There is a time/complexity requirement on the assignment.
I can do that in n*log(m) time where n is the size of the first subset and m is the size of the second subset by doing the following:
for(auto it = subset1.begin(), it != subset1.end(), it++){
if(find(subset2.begin(), subset2.end(), *it))
result.insert(*it);
}
This fails the time requirement, which says worst case linear, but average better than linear.
I found the following question here and I find the hashtable approach interesting. However, I fear that the creation of the hashtable might incur too much overhead. The class contained in the sets looks something like this:
class containedInSets {
//methods
private:
vector<string> member1;
SomeObject member2;
int member3;
}
I have no control over the SomeObject class, and therefore cannot specify a hash function for it. I'd have to hash the pointer. Furthermore, the vector may grow quite (in the thousands of entries).
What is the quickest way of doing this?
Your code is not O(n log(m)) but O(n * m).
std::find(subset2.begin(), subset2.end(), *it) is linear, but std::set has methods find and count which are in O(log(n)) (they do a binary search).
So you can simply do:
for (const auto& e : subset1) {
if (subset2.count(e) != 0) {
result.insert(e);
}
}
Which has complexity of n*log(m) instead of your n * m.
I am trying to solve the following problem: Numbers are being inserted into a container. Each time a number is inserted I need to know how many elements are in the container that are greater than or equal to the current number being inserted. I believe both operations can be done in logarithmic complexity.
My question:
Are there standard containers in a C++ library that can solve the problem?
I know that std::multiset can insert elements in logarithmic time, but how can you query it? Or should I implement a data structure (e.x. a binary search tree) to solve it?
Great question. I do not think there is anything in STL which would suit your needs (provided you MUST have logarithmic times). I think the best solution then, as aschepler says in comments, is to implement a RB tree. You may have a look at STL source code, particularly on stl_tree.h to see whether you could use bits of it.
Better still, look at : (Rank Tree in C++)
Which contains link to implementation:
(http://code.google.com/p/options/downloads/list)
You should use a multiset for logarithmic complexity, yes. But computing the distance is the problem, as set/map iterators are Bidirectional, not RandomAccess, std::distance has an O(n) complexity on them:
multiset<int> my_set;
...
auto it = my_map.lower_bound(3);
size_t count_inserted = distance(it, my_set.end()) // this is definitely O(n)
my_map.insert(make_pair(3);
Your complexity-issue is complicated. Here is a full analysis:
If you want a O(log(n)) complexity for each insertion, you need a sorted structure as a set. If you want the structure to not reallocate or move items when adding a new item, the insertion point distance computation will be O(n). If know the insertion size in advance, you do not need logarithmic insertion time in a sorted container. You can insert all the items then sort, it is as much O(n.log(n)) as n * O(log(n)) insertions in a set.
The only alternative is to use a dedicated container like a weighted RB-tree. Depending on your problem this may be the solution, or something really overkill.
Use multiset and distance, you are O(n.log(n)) on insertion (yes, n insertions * log(n) insertion time for each one of them), O(n.n) on distance computation, but computing distances is very fast.
If you know the inserted data size (n) in advance : Use a vector, fill it, sort it, return your distances, you are O(n.log(n)), and it is easy to code.
If you do not know n in advance, your n is likely huge, each item is memory-heavy so you can not have O(n.log(n)) reallocation : then you have time to re-encode or re-use some non-standard code, you really have to meet these complexity expectations, use a dedicated container. Also consider using a database, you will probably have issues maintaining this in memory.
Here's a quick way using Policy-Based Data Structures in C++:
There exists something called as an Ordered Set, which lets you insert/remove elements in O(logN) time (and pretty much all other functions that std::set has to offer). It also gives 2 more features: Find the Kth element and **find the rank of the Xth element. The problem is that this doesn't allow duplicates :(
No Worries though! We will map duplicates with a separate index/priority, and define a new structure (call it Ordered Multiset)! I've attached my implementation below for reference.
Finally, every time you want to find the no of elements greater than say x, call the function upper_bound (No of elements less than or equal to x) and subtract this number from the size of your Ordered Multiset!
Note: PBDS use a lot of memory, so that is a constraint, I'd suggest using a Binary Search Tree or a Fenwick Tree.
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
struct ordered_multiset { // multiset supporting duplicating values in set
int len = 0;
const int ADD = 1000010;
const int MAXVAL = 1000000010;
unordered_map<int, int> mp; // hash = 96814
tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> T;
ordered_multiset() { len = 0; T.clear(), mp.clear(); }
inline void insert(int x){
len++, x += MAXVAL;
int c = mp[x]++;
T.insert((x * ADD) + c); }
inline void erase(int x){
x += MAXVAL;
int c = mp[x];
if(c) {
c--, mp[x]--, len--;
T.erase((x*ADD) + c); } }
inline int kth(int k){ // 1-based index, returns the
if(k<1 || k>len) return -1; // K'th element in the treap,
auto it = T.find_by_order(--k); // -1 if none exists
return ((*it)/ADD) - MAXVAL; }
inline int lower_bound(int x){ // Count of value <x in treap
x += MAXVAL;
int c = mp[x];
return (T.order_of_key((x*ADD)+c)); }
inline int upper_bound(int x){ // Count of value <=x in treap
x += MAXVAL;
int c = mp[x];
return (T.order_of_key((x*ADD)+c)); }
inline int size() { return len; } // Number of elements in treap
};
Usage:
ordered_multiset s;
for(int i=0; i<n; i++) {
int x; cin>>x;
s.insert(x);
int ctr = s.size() - s.upper_bound(x);
cout<<ctr<<" ";
}
Input (n = 6) : 10 1 3 3 2
Output : 0 1 1 1 3
Time Complexity : O(log n) per query/insert
References : mochow13's GitHub
Sounds like a case for count_if - although I admit this doesn't solve it at logarithmic complexity, that would require a sorted type.
vector<int> v = { 1, 2, 3, 4, 5 };
int some_value = 3;
int count = count_if(v.begin(), v.end(), [some_value](int n) { return n > some_value; } );
Edit done to fix syntactic problems with lambda function
If the whole range of numbers is sufficiently small (on the order of a few million), this problem can be solved relatively easily using a Fenwick tree.
Although Fenwick trees are not part of the STL, they are both very easy to implement and time efficient. The time complexity is O(log N) for both updates and queries and the constant factors are low.
You mention in a comment on another question, that you needed this for a contest. Fenwick trees are very popular tools in competitive programming and are often useful.
I was asked an interview question to find the number of distinct absolute values among the elements of the array. I came up with the following solution (in C++) but the interviewer was not happy with the code's run time efficiency.
I will appreciate pointers as to how I can improve the run time efficiency of this code?
Also how do I calculate the efficiency of the code below? The for loop executes A.size() times. However I am not sure about the efficiency of STL std::find (In the worse case it could be O(n) so that makes this code O(n²) ?
Code is:
int countAbsoluteDistinct ( const std::vector<int> &A ) {
using namespace std;
list<int> x;
vector<int>::const_iterator it;
for(it = A.begin();it < A.end();it++)
if(find(x.begin(),x.end(),abs(*it)) == x.end())
x.push_back(abs(*it));
return x.size();
}
To propose alternative code to the set code.
Note that we don't want to alter the caller's vector, we take by value. It's better to let the compiler copy for us than make our own. If it's ok to destroy their value we can take by non-const reference.
#include <vector>
#include <algorithm>
#include <iterator>
#include <cstdlib>
using namespace std;
int count_distinct_abs(vector<int> v)
{
transform(v.begin(), v.end(), v.begin(), abs); // O(n) where n = distance(v.end(), v.begin())
sort(v.begin(), v.end()); // Average case O(n log n), worst case O(n^2) (usually implemented as quicksort.
// To guarantee worst case O(n log n) replace with make_heap, then sort_heap.
// Unique will take a sorted range, and move things around to get duplicated
// items to the back and returns an iterator to the end of the unique section of the range
auto unique_end = unique(v.begin(), v.end()); // Again n comparisons
return distance(v.begin(), unique_end); // Constant time for random access iterators (like vector's)
}
The advantage here is that we only allocate/copy once if we decide to take by value, and the rest is all done in-place while still giving you an average complexity of O(n log n) on the size of v.
std::find() is linear (O(n)). I'd use a sorted associative container to handle this, specifically std::set.
#include <vector>
#include <set>
using namespace std;
int distict_abs(const vector<int>& v)
{
std::set<int> distinct_container;
for(auto curr_int = v.begin(), end = v.end(); // no need to call v.end() multiple times
curr_int != end;
++curr_int)
{
// std::set only allows single entries
// since that is what we want, we don't care that this fails
// if the second (or more) of the same value is attempted to
// be inserted.
distinct_container.insert(abs(*curr_int));
}
return distinct_container.size();
}
There is still some runtime penalty with this approach. Using a separate container incurs the cost of dynamic allocations as the container size increases. You could do this in place and not occur this penalty, however with code at this level its sometimes better to be clear and explicit and let the optimizer (in the compiler) do its work.
Yes, this will be O(N2) -- you'll end up with a linear search for each element.
A couple of reasonably obvious alternatives would be to use an std::set or std::unordered_set. If you don't have C++0x, you can replace std::unordered_set with tr1::unordered_set or boost::unordered_set.
Each insertion in an std::set is O(log N), so your overall complexity is O(N log N).
With unordered_set, each insertion has constant (expected) complexity, giving linear complexity overall.
Basically, replace your std::list with a std::set. This gives you O(log(set.size())) searches + O(1) insertions, if you do things properly. Also, for efficiency, it makes sense to cache the result of abs(*it), although this will have only a minimal (negligible) effect. The efficiency of this method is about as good as you can get it, without using a really nice hash (std::set uses bin-trees) or more information about the values in the vector.
Since I was not happy with the previous answer here is mine today. Your intial question does not mention how big your vector is. Suppose your std::vector<> is extremely large and have very few duplicates (why not?). This means that using another container (eg. std::set<>) will basically duplicate your memory consumption. Why would you do that since your goal is simply to count non duplicate.
I like #Flame answer, but I was not really happy with the call to std::unique. You've spent lots of time carefully sorting your vector and then simply discard the sorted array while you could be re-using it afterward.
I could not find anything really elegant in the STD library, so here is my proposal (a mixture of std::transform + std::abs + std::sort, but without touching the sorted array afterward).
// count the number of distinct absolute values among the elements of the sorted container
template<class ForwardIt>
typename std::iterator_traits<ForwardIt>::difference_type
count_unique(ForwardIt first, ForwardIt last)
{
if (first == last)
return 0;
typename std::iterator_traits<ForwardIt>::difference_type
count = 1;
ForwardIt previous = first;
while (++first != last) {
if (!(*previous == *first) ) ++count;
++previous;
}
return count;
}
Bonus point is works with forward iterator:
#include <iostream>
#include <list>
int main()
{
std::list<int> nums {1, 3, 3, 3, 5, 5, 7,8};
std::cout << count_unique( std::begin(nums), std::end(nums) ) << std::endl;
const int array[] = { 0,0,0,1,2,3,3,3,4,4,4,4};
const int n = sizeof array / sizeof * array;
std::cout << count_unique( array, array + n ) << std::endl;
return 0;
}
Two points.
std::list is very bad for search. Each search is O(n).
Use std::set. Insert is logarithmic, it removes duplicate and is sorted. Insert every value O(n log n) then use set::size to find how many values.
EDIT:
To answer part 2 of your question, the C++ standard mandates the worst case for operations on containers and algorithms.
Find: Since you are using the free function version of find which takes iterators, it cannot assume anything about the passed in sequence, it cannot assume that the range is sorted, so it must traverse every item until it finds a match, which is O(n).
If you are using set::find on the other hand, this member find can utilize the structure of the set, and it's performance is required to be O(log N) where N is the size of the set.
To answer your second question first, yes the code is O(n^2) because the complexity of find is O(n).
You have options to improve it. If the range of numbers is low you can just set up a large enough array and increment counts while iterating over the source data. If the range is larger but sparse, you can use a hash table of some sort to do the counting. Both of these options are linear complexity.
Otherwise, I would do one iteration to take the abs value of each item, then sort them, and then you can do the aggregation in a single additional pass. The complexity here is n log(n) for the sort. The other passes don't matter for complexity.
I think a std::map could also be interesting:
int absoluteDistinct(const vector<int> &A)
{
map<int, char> my_map;
for (vector<int>::const_iterator it = A.begin(); it != A.end(); it++)
{
my_map[abs(*it)] = 0;
}
return my_map.size();
}
As #Jerry said, to improve a little on the theme of most of the other answers, instead of using a std::map or std::set you could use a std::unordered_map or std::unordered_set (or the boost equivalent).
This would reduce the runtimes down from O(n lg n) or O(n).
Another possibility, depending on the range of the data given, you might be able to do a variant of a radix sort, though there's nothing in the question that immediately suggests this.
Sort the list with a Radix style sort for O(n)ish efficiency. Compare adjacent values.
The best way is to customize the quicksort algorithm such that when we are partitioning whenever we get two equal element then overwrite the second duplicate with last element in the range and then reduce the range. This will ensure you will not process duplicate elements twice. Also after quick sort is done the range of the element is answer
Complexity is still O(n*Lg-n) BUT this should save atleast two passes over the array.
Also savings are proportional to % of duplicates. Imagine if they twist original questoin with, 'say 90% of the elements are duplicate' ...
One more approach :
Space efficient : Use hash map .
O(logN)*O(n) for insert and just keep the count of number of elements successfully inserted.
Time efficient : Use hash table O(n) for insert and just keep the count of number of elements successfully inserted.
You have nested loops in your code. If you will scan each element over the whole array it will give you O(n^2) time complexity which is not acceptable in most of the scenarios. That was the reason the Merge Sort and Quick sort algorithms came up to save processing cycles and machine efforts. I will suggest you to go through the suggested links and redesign your program.
I have a data structure like this:
struct X {
float value;
int id;
};
a vector of those (size N (think 100000), sorted by value (stays constant during the execution of the program):
std::vector<X> values;
Now, I want to write a function
void subvector(std::vector<X> const& values,
std::vector<int> const& ids,
std::vector<X>& out /*,
helper data here */);
that fills the out parameter with a sorted subset of values, given by the passed ids (size M < N (about 0.8 times N)), fast (memory is not an issue, and this will be done repeatedly, so building lookuptables (the helper data from the function parameters) or something else that is done only once is entirely ok).
My solution so far:
Build lookuptable lut containing id -> offset in values (preparation, so constant runtime)
create std::vector<X> tmp, size N, filled with invalid ids (linear in N)
for each id, copy values[lut[id]] to tmp[lut[id]] (linear in M)
loop over tmp, copying items to out (linear in N)
this is linear in N (as it's bigger than M), but the temporary variable and repeated copying bugs me. Is there a way to do it quicker than this? Note that M will be close to N, so things that are O(M log N) are unfavourable.
Edit: http://ideone.com/xR8Vp is a sample implementation of mentioned algorithm, to make the desired output clear and prove that it's doable in linear time - the question is about the possibility of avoiding the temporary variable or speeding it up in some other way, something that is not linear is not faster :).
An alternative approach you could try is to use a hash table instead of a vector to look up ids in:
void subvector(std::vector<X> const& values,
std::unordered_set<int> const& ids,
std::vector<X>& out) {
out.clear();
out.reserve(ids.size());
for(std::vector<X>::const_iterator i = values.begin(); i != values.end(); ++i) {
if(ids.find(i->id) != ids.end()) {
out.push_back(*i);
}
}
}
This runs in linear time since unordered_set::find is constant expected time (assuming that we have no problems hashing ints). However I suspect it might not be as fast in practice as the approach you described initially using vectors.
Since your vector is sorted, and you want a subset of it sorted the same way, I assume we can just slice out the chunk you want without rearranging it.
Why not just use find_if() twice. Once to find the start of the range you want and once to find the end of the range. This will give you the start and end iterators of the sub vector. Construct a new vector using those iterators. One of the vector constructor overloads takes two iterators.
That or the partition algorithm should work.
If I understood your problem correctly, you actually try to create a linear time sorting algorithm (subject to the input size of numbers M).
That is NOT possible.
Your current approach is to have a sorted list of possible values.
This takes linear time to the number of possible values N (theoretically, given that the map search takes O(1) time).
The best you could do, is to sort the values (you found from the map) with a quick sorting method (O(MlogM) f.e. quicksort, mergesort etc) for small values of M and maybe do that linear search for bigger values of M.
For example, if N is 100000 and M is 100 it is much faster to just use a sorting algorithm.
I hope you can understand what I say. If you still have questions I will try to answer them :)
edit: (comment)
I will further explain what I mean.
Say you know that your numbers will range from 1 to 100.
You have them sorted somewhere (actually they are "naturally" sorted) and you want to get a subset of them in sorted form.
If it would be possible to do it faster than O(N) or O(MlogM), sorting algorithms would just use this method to sort.
F.e. by having the set of numbers {5,10,3,8,9,1,7}, knowing that they are a subset of the sorted set of numbers {1,2,3,4,5,6,7,8,9,10} you still can't sort them faster than O(N) (N = 10) or O(MlogM) (M = 7).