Get top 5 algorithm from a container? - c++

I have a class(object), User. This user has 2 private attributes, "name" and "popularity". I store the objects into a vector (container).
From the container, I need to find the top 5 most popular user, how do I do that? (I have an ugly code, I will post here, if you have a better approach, please let me know. Feel free to use other container, if you think vector is not a good choice, but please use only: map or multimap, list, vector or array, because I only know how to use these.) My current code is:
int top5 = 0, top4 = 0, top3 = 0, top2 = 0, top1 = 0;
vector<User>::iterator it;
for (it = user.begin(); it != user.end(); ++it)
{
if( it->getPopularity() > top5){
if(it->getPopularity() > top4){
if(it->getPopularity() > top3){
if(it->getPopularity() > top2){
if(it->getPopularity() > top1){
top1 = it->getPopularity();
continue;
} else {
top2 = it->getPopularity();
continue;
}
} else {
top3 = it->getPopularity();
continue;
}
}
} else {
top4 = it->getPopularity();
continue;
}
} else {
top5 = it->getPopularity();
continue;
}
}
I know the codes is ugly and might be prone to error, thus if you have better codes, please do share with us (us == cpp newbie). Thanks

You can use the std::partial_sort algorithm to sort your vector so that the first five elements are sorted and the rest remains unsorted. Something like this (untested code):
bool compareByPopularity( User a, User b ) {
return a.GetPopularity() > b.GetPopularity();
}
vector<Users> getMostPopularUsers( const vector<User> &users, int num ) {
if ( users.size() <= num ) {
sort( users.begin(), users.end(), compareByPopularity );
} else {
partial_sort( users.begin(), users.begin() + num, users.end(),
compareByPopularity );
}
return vector<Users>( users.begin(), users.begin() + num );
}

Why don't you sort (std::sort or your own implementation of Quick Sort) the vector based on popularity and take the first 5 values ?
Example:
bool UserCompare(User a, User b) { return a.getPopularity() > b.getPopularity(); }
...
std::sort(user.begin(), user.end(), UserCompare);
// Print first 5 users

If you just want top 5 popular uses, then use std::partial_sort().
class User
{
private:
string name_m;
int popularity_m;
public:
User(const string& name, int popularity) : name_m(name), popularity_m(popularity) { }
friend ostream& operator<<(ostream& os, const User& user)
{
return os << "name:" << user.name_m << "|popularity:" << user.popularity_m << "\n";
return os;
}
int Popularity() const
{
return popularity_m;
}
};
bool Compare(const User& lhs, const User& rhs)
{
return lhs.Popularity() > rhs.Popularity();
}
int main()
{
// c++0x. ignore if you don't want it.
auto compare = [](const User& lhs, const User& rhs) -> bool
{ return lhs.Popularity() > rhs.Popularity(); };
partial_sort(users.begin(), users.begin() + 5, users.end(), Compare);
copy(users.begin(), users.begin() + 5, ostream_iterator<User>(std::cout, "\n"));
}

First off, cache that it->getPopularity() so you don't have to keep repeating it.
Secondly (and this is much more important): Your algorithm is flawed. When you find a new top1 you have to push the old top1 down to the #2 slot before you save the new top1, but before you do that you have to push the old top2 down to the #3 slot, etc. And that is just for a new top1. You are going to have to do something similar for a new top2, a new top3, etc. The only one you can paste in without worrying about pushing things down the list is when you get a new top5. The correct algorithm is hairy. That said, the correct algorithm is much easier to implement when your topN is an array rather than a bunch of separate values.
Thirdly (and this is even more important than the second point): You shouldn't care about performance, at least not initially. The easy way to do this is to sort the entire list and pluck off the first five off the top. If this suboptimal but simple algorithm doesn't affect your performance, done. Don't bother with the ugly but fast first N algorithm unless performance mandates that you toss the simple solution out the window.
Finally (and this is the most important point of all): That fast first N algorithm is only fast when the number of elements in the list is much, much larger than five. The default sort algorithm is pretty dang fast. It has to be wasting a lot of time sorting the dozens / hundreds of items you don't care about before a pushdown first N algorithm becomes advantageous. In other words, that pushdown insertion sort algorithm may well be a case of premature disoptimization.

Sort your objects, maybe with the library if this is allowed, and then simply selecte the first 5 element. If your container gets too big you could probably use a std::list for the job.
Edit : #itsik you beat me to the sec :)

Do this pseudo code.
Declare top5 as an array of int[5] // or use a min-heap
Initialize top5 as 5 -INF
For each element A
if A < top5[4] // or A < root-of-top5
Remove top5[4] from top5 // or pop min element from heap
Insert A to top // or insert A to the heap

Well, I advise you improve your code by using an array or list or vector to store the top five, like this
struct TopRecord
{
int index;
int pop;
} Top5[5];
for(int i = 0; i<5; i++)
{
Top5[i].index = -1;
// Set pop to a value low enough
Top5[i].pop = -1;
}
for(int i = 0; i< users.size(); i++)
{
int currentpop = i->getPopularity()
int currentindex = i;
int j = 0;
int temp;
while(j < 5 && Top5[j].pop < currentpop)
{
temp = Top5[j].pop;
Top[j].pop = currentpop;
currentpop = temp;
temp = Top5[j].index;
Top[j].index = currentindex;
currentindex = temp;
j++;
}
}

You also may consider using Randomized Select if Your aim is performance, since originally Randomized Select is good enough for ordered statistics and runs in linear time, You just need to run it 5 times. Or to use partial_sort solution provided above, either way counts, depends on Your aim.

Related

fast way to compare two vector containing strings

I have a vector of strings I that pass to my function and I need to compare it with some pre-defined values. What is the fastest way to do this?
The following code snippet shows what I need to do (This is how I am doing it, but what is the fastest way of doing this):
bool compare(vector<string> input1,vector<string> input2)
{
if(input1.size() != input2.size()
{
return false;
}
for(int i=0;i<input1.siz();i++)
{
if(input1[i] != input2[i])
{
return false;
}
}
return true;
}
int compare(vector<string> inputData)
{
if (compare(inputData,{"Apple","Orange","three"}))
{
return 129;
}
if (compare(inputData,{"A","B","CCC"}))
{
return 189;
}
if (compare(inputData,{"s","O","quick"}))
{
return 126;
}
if (compare(inputData,{"Apple","O123","three","four","five","six"}))
{
return 876;
}
if (compare(inputData,{"Apple","iuyt","asde","qwe","asdr"}))
{
return 234;
}
return 0;
}
Edit1
Can I compare two vector like this:
if(inputData=={"Apple","Orange","three"})
{
return 129;
}
You are asking what is the fastest way to do this, and you are indicating that you are comparing against a set of fixed and known strings. I would argue that you would probably have to implement it as a kind of state machine. Not that this is very beautiful...
if (inputData.size() != 3) return 0;
if (inputData[0].size() == 0) return 0;
const char inputData_0_0 = inputData[0][0];
if (inputData_0_0 == 'A') {
// possibly "Apple" or "A"
...
} else if (inputData_0_0 == 's') {
// possibly "s"
...
} else {
return 0;
}
The weakness of your approach is its linearity. You want a binary search for teh speedz.
By utilising the sortedness of a map, the binaryness of finding in one, and the fact that equivalence between vectors is already defined for you (no need for that first compare function!), you can do this quite easily:
std::map<std::vector<std::string>, int> lookup{
{{"Apple","Orange","three"}, 129},
{{"A","B","CCC"}, 189},
// ...
};
int compare(const std::vector<std::string>& inputData)
{
auto it = lookup.find(inputData);
if (it != lookup.end())
return it->second;
else
return 0;
}
Note also the reference passing for extra teh speedz.
(I haven't tested this for exact syntax-correctness, but you get the idea.)
However! As always, we need to be context-aware in our designs. This sort of approach is more useful at larger scale. At the moment you only have a few options, so the addition of some dynamic allocation and sorting and all that jazz may actually slow things down. Ultimately, you will want to take my solution, and your solution, and measure the results for typical inputs and whatnot.
Once you've done that, if you still need more speed for some reason, consider looking at ways to reduce the dynamic allocations inherent in both the vectors and the strings themselves.
To answer your follow-up question: almost; you do need to specify the type:
// new code is here
// ||||||||||||||||||||||||
if (inputData == std::vector<std::string>{"Apple","Orange","three"})
{
return 129;
}
As explored above, though, let std::map::find do this for you instead. It's better at it.
One key to efficiency is eliminating needless allocation.
Thus, it becomes:
bool compare(
std::vector<std::string> const& a,
std::initializer_list<const char*> b
) noexcept {
return std::equal(begin(a), end(a), begin(b), end(b));
}
Alternatively, make them static const, and accept the slight overhead.
As an aside, using C++17 std::string_view (look at boost), C++20 std::span (look for the Guideline support library (GSL)) also allows a nicer alternative:
bool compare(std::span<std::string> a, std::span<std::string_view> b) noexcept {
return a == b;
}
The other is minimizing the number of comparisons. You can either use hashing, binary search, or manual ordering of comparisons.
Unfortunately, transparent comparators are a C++14 thing, so you cannot use std::map.
If you want a fast way to do it where the vectors to compare to are not known in advance, but are reused so can have a little initial run-time overhead, you can build a tree structure similar to the compile time version Dirk Herrmann has. This will run in O(n) by just iterating over the input and following a tree.
In the simplest case, you might build a tree for each letter/element. A partial implementation could be:
typedef std::vector<std::string> Vector;
typedef Vector::const_iterator Iterator;
typedef std::string::const_iterator StrIterator;
struct Node
{
std::unique_ptr<Node> children[256];
std::unique_ptr<Node> new_str_child;
int result;
bool is_result;
};
Node root;
int compare(Iterator vec_it, Iterator vec_end, StrIterator str_it, StrIterator str_end, const Node *node);
int compare(const Vector &input)
{
return compare(input.begin(), input.end(), input.front().begin(), input.front().end(), &root);
}
int compare(Iterator vec_it, Iterator vec_end, StrIterator str_it, StrIterator str_end, const Node *node)
{
if (str_it != str_end)
{
// Check next character
auto next_child = node->children[(unsigned char)*str_it].get();
if (next_child)
return compare(vec_it, vec_end, str_it + 1, str_end, next_child);
else return -1; // No string matched
}
// At end of input string
++vec_it;
if (vec_it != vec_end)
{
auto next_child = node->new_str_child.get();
if (next_child)
return compare(vec_it, vec_end, vec_it->begin(), vec_it->end(), next_child);
else return -1; // Have another string, but not in tree
}
// At end of input vector
if (node->is_result)
return node->result; // Got a match
else return -1; // Run out of input, but all possible matches were longer
}
Which can also be done without recursion. For use cases like yours you will find most nodes only have a single success value, so you can collapse those into prefix substrings, to use the OP example:
"A"
|-"pple" - new vector - "O" - "range" - new vector - "three" - ret 129
| |- "i" - "uyt" - new vector - "asde" ... - ret 234
| |- "0" - "123" - new vector - "three" ... - ret 876
|- new vector "B" - new vector - "CCC" - ret 189
"s" - new vector "O" - new vector "quick" - ret 126
you could make use of std::equal function like below :
bool compare(vector<string> input1,vector<string> input2)
{
if(input1.size() != input2.size()
{
return false;
}
return std::equal(input1.begin(), input2.end(), input2.begin())
}
Can I compare two vector like this
The answer is No, you need compare a vector with another vector, like this:
vector<string>data = {"ab", "cd", "ef"};
if(data == vector<string>{"ab", "cd", "efg"})
cout << "Equal" << endl;
else
cout << "Not Equal" << endl;
What is the fastest way to do this?
I'm not an expert of asymptotic analysis but:
Using the relational operator equality (==) you have a shortcut to compare two vectors, first validating the size and, second, each element on them. This way provide a linear execution (T(n), where n is the size of vector) which compare each item of the vector, but each string must be compared and, generally, it is another linear comparison (T(m), where m is the size of the string).
Suppose that each string has de same size (m) and you have a vector of size n, each comparison could have a behavior of T(nm).
So:
if you want a shortcut to compare two vector you can use the
relational operator equality.
If you want an program which perform a fast comparison you should look for some algorithm for compare strings.

A* and N-Puzzle optimization

I am writing a solver for the N-Puzzle (see http://en.wikipedia.org/wiki/Fifteen_puzzle)
Right now I am using a unordered_map to store hash values of the puzzle board,
and manhattan distance as the heuristic for the algorithm, which is a plain DFS.
so I have
auto pred = [](Node * lhs, Node * rhs){ return lhs->manhattanCost_ < rhs->manhattanCost_; };
std::multiset<Node *, decltype(pred)> frontier(pred);
std::vector<Node *> explored; // holds nodes we have already explored
std::tr1::unordered_set<unsigned> frontierHashTable;
std::tr1::unordered_set<unsigned> exploredHashTable;
This works great for n = 2 and 3.
However, its really hit and miss for n=4 and above. (stl unable to allocate memory for a new node)
I also suspect that I am getting hash collisions in the unordered_set
unsigned makeHash(const Node & pNode)
{
unsigned int b = 378551;
unsigned int a = 63689;
unsigned int hash = 0;
for(std::size_t i = 0; i < pNode.data_.size(); i++)
{
hash = hash * a + pNode.data_[i];
a = a * b;
}
return hash;
}
16! = 2 × 10^13 (possible arrangements)
2^32 = 4 x 10^9 (possible hash values in a 32 bit hash)
My question is how can I optimize my code to solve for n=4 and n=5?
I know from here
http://kociemba.org/fifteen/fifteensolver.html
http://www.ic-net.or.jp/home/takaken/e/15pz/index.html
that n=4 is possible in less than a second on average.
edit:
The algorithm itself is here:
bool NPuzzle::aStarSearch()
{
auto pred = [](Node * lhs, Node * rhs){ return lhs->manhattanCost_ < rhs->manhattanCost_; };
std::multiset<Node *, decltype(pred)> frontier(pred);
std::vector<Node *> explored; // holds nodes we have already explored
std::tr1::unordered_set<unsigned> frontierHashTable;
std::tr1::unordered_set<unsigned> exploredHashTable;
// if we are in the solved position in the first place, return true
if(initial_ == target_)
{
current_ = initial_;
return true;
}
frontier.insert(new Node(initial_)); // we are going to delete everything from the frontier later..
for(;;)
{
if(frontier.empty())
{
std::cout << "depth first search " << "cant solve!" << std::endl;
return false;
}
// remove a node from the frontier, and place it into the explored set
Node * pLeaf = *frontier.begin();
frontier.erase(frontier.begin());
explored.push_back(pLeaf);
// do the same for the hash table
unsigned hashValue = makeHash(*pLeaf);
frontierHashTable.erase(hashValue);
exploredHashTable.insert(hashValue);
std::vector<Node *> children = pLeaf->genChildren();
for( auto it = children.begin(); it != children.end(); ++it)
{
unsigned childHash = makeHash(**it);
if(inFrontierOrExplored(frontierHashTable, exploredHashTable, childHash))
{
delete *it;
}
else
{
if(**it == target_)
{
explored.push_back(*it);
current_ = **it;
// delete everything else in children
for( auto it2 = ++it; it2 != children.end(); ++it2)
delete * it2;
// delete everything in the frontier
for( auto it = frontier.begin(); it != frontier.end(); ++it)
delete *it;
// delete everything in explored
explored_.swap(explored);
for( auto it = explored.begin(); it != explored.end(); ++it)
delete *it;
return true;
}
else
{
frontier.insert(*it);
frontierHashTable.insert(childHash);
}
}
}
}
}
Since this is homework I will suggest some strategies you might try.
First, try using valgrind or a similar tool to check for memory leaks. You may have some memory leaks if you don't delete everything you new.
Second, calculate a bound on the number of nodes that should be explored. Keep track of the number of nodes you do explore. If you pass the bound, you might not be detecting cycles properly.
Third, try the algorithm with depth first search instead of A*. Its memory requirements should be linear in the depth of the tree and it should just be a matter of changing the sort ordering (pred). If DFS works, your A* search may be exploring too many nodes or your memory structures might be too inefficient. If DFS doesn't work, again it might be a problem with cycles.
Fourth, try more compact memory structures. For example, std::multiset does what you want but std::priority_queue with a std::deque may take up less memory. There are other changes you could try and see if they improve things.
First i would recommend cantor expansion, which you can use as the hashing method. It's 1-to-1, i.e. the 16! possible arrangements would be hashed into 0 ~ 16! - 1.
And then i would implement map by my self, as you may know, std is not efficient enough for computation. map is actually a Binary Search Tree, i would recommend Size Balanced Tree, or you can use AVL tree.
And just for record, directly use bool hash[] & big prime may also receive good result.
Then the most important thing - the A* function, like what's in the first of your link, you may try variety of A* function and find the best one.
You are only using the heuristic function to order the multiset. You should use the min(g(n) + f(n)) i.e. the min(path length + heuristic) to order your frontier.
Here the problem is, you are picking the one with the least heuristic, which may not be the correct "next child" to pick.
I believe this is what is causing your calculation to explode.

How to implement C++ dictionary data structure without using STL

I have a project that I am doing, the main objective is to load a list of words (and lots of them 15k+) into a data structure and then do a search on that structure. I did a little research and as far as I can tell a hash table would be best for this (correct me if I am wrong, I looked into tries as well)
Here's the tricky part: I cannot use any STL's for this project. So as far as I can tell I am going to have to write my own hash table class or find one that pretty much works. I understand how has tables work on a basic level but I am not sure I know enough to write a whole one by myself.
I looked around Google and I could not find any suitable sample code.
My question is does anyone know how to do this in c++ and/or where I can find some code to start off with. I need 3 basic functions for the table: insert, search, remove.
Things to remember while you think about this:
The Number 1 Concern is SPEED! this needs to be lighting fast, no concern for system resources. From the reading that I have done, a hash table can do better than O(log n) Consider mutithreading?
Cannot use STL!
I think, sorted array of strings + binary search should be pretty efficient.
std::unordered_map is not STL
http://www.cs.auckland.ac.nz/software/AlgAnim/hash_tables.html
Not entirely clear on all of the restrictions, but assuming you cannot use anything from std, you could write a simple class like the one below to do the job. We will use an array of buckets to store the data, then use a hash function to turn a string into a number in the range 0...MAX_ELEMENTS. each bucket will hold a linked list of strings, so you can retrieve information again. Typically o(1) insertion and find.
Note that for a more effective solution, you may wish to use a vector rather than a fixed length array as I have gone for. There is also minimal error checking and other improvements, but this should get you started.
NOTE you will need to implement your own string hashing function, you can find plenty of these on the net.
class dictionary
{
struct data
{
char* word = nullptr;
data* next = nullptr;
~data()
{
delete [] word;
}
};
public:
const unsigned int MAX_BUCKETS;
dictionary(unsigned int maxBuckets = 1024)
: MAX_BUCKETS(maxBuckets)
, words(new data*[MAX_BUCKETS])
{
memset(words, 0, sizeof(data*) * MAX_BUCKETS);
}
~dictionary()
{
for (int i = 0; i < MAX_BUCKETS; ++i)
delete words[i];
delete [] words;
}
void insert(const char* word)
{
const auto hash_index = hash(word);
auto& d = words[hash_index];
if (d == nullptr)
{
d = new data;
copy_string(d, word);
}
else
{
while (d->next != nullptr)
{
d = d->next;
}
d->next = new data;
copy_string(d->next, word);
}
}
void copy_string(data* d, const char* word)
{
const auto word_length = strlen(word)+1;
d->word = new char[word_length];
strcpy(d->word, word);
printf("%s\n", d->word);
}
const char* find(const char* word) const
{
const auto hash_index = hash(word);
auto& d = words[hash_index];
if (d == nullptr)
{
return nullptr;
}
while (d != nullptr)
{
printf("checking %s with %s\n", word, d->word);
if (strcmp(d->word, word) == 0)
return d->word;
d = d->next;
}
return nullptr;
}
private:
unsigned int hash(const char* word) const
{
// :TODO: write your own hash function here
const unsigned int num = 0; // :TODO:
return num % MAX_BUCKETS;
}
data** words;
};
http://wikipedia-clustering.speedblue.org/trie.php
Seems the above link is down at the moment.
Alternative link:
https://web.archive.org/web/20160426224744/http://wikipedia-clustering.speedblue.org/trie.php
Source Code: https://web.archive.org/web/20160426224744/http://wikipedia-clustering.speedblue.org/download/libTrie-0.1.tgz

Shifting elements in an array C++

I've developed a method called "rotate" to my stack object class. What I did was that if the stack contains elements: {0,2,3,4,5,6,7} I would needed to rotate the elements forwards and backwards.
Where if i need to rotate forwards by 2 elements, then we would have, {3,4,5,6,7,0,2} in the array. And if I need to rotate backwards, or -3 elements, then, looking at the original array it would be, {5,6,7,0,2,3,4}
So the method that I have developed works fine. Its just terribly ineffecient IMO. I was wondering if I could wrap the array around by using the mod operator? Or if their is useless code hangin' around that I havent realized yet, and so on.
I guess my question is, How can i simplify this method? e.g. using less code. :-)
void stack::rotate(int r)
{
int i = 0;
while ( r > 0 ) // rotate postively.
{
front.n = items[top+1].n;
for ( int j = 0; j < bottom; j++ )
{
items[j] = items[j+1];
}
items[count-1].n = front.n;
r--;
}
while ( r < 0 ) // rotate negatively.
{
if ( i == top+1 )
{
front.n = items[top+1].n;
items[top+1].n = items[count-1].n; // switch last with first
}
back.n = items[++i].n; // second element is the new back
items[i].n = front.n;
if ( i == bottom )
{
items[count-1].n = front.n; // last is first
i = 0;
r++;
continue;
}
else
{
front.n = items[++i].n;
items[i].n = back.n;
if ( i == bottom )
{
i = 0;
r++;
continue;
}
}
}
}
Instead of moving all the items in your stack, you could change the definition of 'beginning'. Have an index that represents the first item in the stack, 0 at the start, which you add to and subtract from using modular arithmetic whenever you want to rotate your stack.
Note that if you take this approach you shouldn't give users of your class access to the underlying array (not that you really should anyway...).
Well, as this is an abstraction around an array, you can store the "zero" index as a member of the abstraction, and index into the array based on this abstract notion of the first element. Roughly...
class WrappedArray
{
int length;
int first;
T *array;
T get(int index)
{
return array[(first + index) % length];
}
int rotateForwards()
{
first++;
if (first == length)
first = 0;
}
}
You've gotten a couple of reasonable answers, already, but perhaps one more won't hurt. My first reaction would be to make your stack a wrapper around an std::deque, in which case moving an element from one end to the other is cheap (O(1)).
What you are after here is a circular list.
If you insist on storing items in an array just use top offset and size for access. This approach makes inserting elements after you reached allocated size expensive though (re-allocation, copying). This can be solved by using doubly-linked list (ala std::list) and an iterator, but arbitrary access into the stack will be O(n).
The function rotate below is based on reminders (do you mean this under the 'mod' operation?)
It is also quite efficient.
// Helper function.
// Finds GCD.
// See http://en.wikipedia.org/wiki/Euclidean_algorithm#Implementations
int gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);}
// Number of assignments of elements in algo is
// equal to (items.size() + gcd(items.size(),r)).
void rotate(std::vector<int>& items, int r) {
int size = (int)items.size();
if (size <= 1) return; // nothing to do
r = (r % size + size) % size; // fits r into [0..size)
int num_cycles = gcd(size, r);
for (int first_index = 0; first_index < num_cycles; ++first_index) {
int mem = items[first_index]; // assignment of items elements
int index = (first_index + r) % size, index_prev = first_index;
while (index != first_index) {
items[index_prev] = items[index]; // assignment of items elements
index_prev = index;
index = (index + r) % size;
};
items[index_prev] = mem; // assignment of items elements
}
}
Of course if it is appropriate for you to change data structure as described in other answers, you can obtain more efficient solution.
And now, the usual "it's already in Boost" answer: There is a Boost.CircularBuffer
If for some reason you'd prefer to perform actual physical rotation of array elements, you might find several alternative solutions in "Programming Pearls" by Jon Bentley (Column 2, 2.3 The Power of Primitives). Actually a Web search for Rotating Algorithms 'Programming Pearls' will tell you everything. The literal approach you are using now has very little practical value.
If you'd prefer to try to solve it yourself, it might help to try looking at the problem differently. You see, "rotating an array" is really the same thing as "swapping two unequal parts of an array". Thinking about this problem in the latter terms might lead you to new solutions :)
For example,
Reversal Approach. Reverse the order of the elements in the entire array. Then reverse the two parts independently. You are done.
For example, let's say we want to rotate abcdefg right by 2
abcdefg -> reverse the whole -> gfedcba -> reverse the two parts -> fgabcde
P.S. Slides for that chapter of "Programming Pearls". Note that in Bentley's experiments the above algorithm proves to be quite efficient (among the three tested).
I don't understand what the variables front and back mean, and why you need .n. Anyway, this is the shortest code I know to rotate the elements of an array, which can also be found in Bentley's book.
#include <algorithm>
std::reverse(array , array + r );
std::reverse(array + r, array + size);
std::reverse(array , array + size);

PushFront method for an array C++

I thought i'd post a little of my homework assignment. Im so lost in it. I just have to be really efficient. Without using any stls, boosts and the like. By this post, I was hoping that someone could help me figure it out.
bool stack::pushFront(const int nPushFront)
{
if ( count == maxSize ) // indicates a full array
{
return false;
}
else if ( count <= 0 )
{
count++;
items[top+1].n = nPushFront;
return true;
}
++count;
for ( int i = 0; i < count - 1; i++ )
{
intBackPtr = intFrontPtr;
intBackPtr++;
*intBackPtr = *intFrontPtr;
}
items[top+1].n = nPushFront;
return true;
}
I just cannot figure out for the life of me to do this correctly! I hope im doing this right, what with the pointers and all
int *intFrontPtr = &items[0].n;
int *intBackPtr = &items[capacity-1].n;
Im trying to think of this pushFront method like shifting an array to the right by 'n' units...I can only seem to do that in an array that is full. Can someone out their please help me?
Firstly, I'm not sure why you have the line else if ( count <= 0 ) - the count of items in your stack should never be below 0.
Usually, you would implement a stack not by pushing to the front, but pushing and popping from the back. So rather than moving everything along, as it looks like you're doing, just store a pointer to where the last element is, and insert just after that, and pop from there. When you push, just increment that pointer, and when you pop, decrement it (you don't even have to delete it). If that pointer is at the end of your array, you're full (so you don't even have to store a count value). And if it's at the start, then it's empty.
Edit
If you're after a queue, look into Circular Queues. That's typically how you'd implement one in an array. Alternatively, rather than using an array, try a Linked List - that lets it be arbitrarily big (the only limit is your computer's memory).
You don't need any pointers to shift an array. Just use simple for statement:
int *a; // Your array
int count; // Elements count in array
int length; // Length of array (maxSize)
bool pushFront(const int nPushFront)
{
if (count == length) return false;
for (int i = count - 1; i >= 0; --i)
Swap(a[i], a[i + 1]);
a[0] = nPushFront; ++count;
return true;
}
Without doing your homework for you let me see if I can give you some hints. Implementing a deque (double ended queue) is really quite easy if you can get your head around a few concepts.
Firstly, it is key to note that since we will be popping off the front and/or back in order to efficiently code an algorithm which uses contiguous storage we need to be able to pop front/back without shifting the entire array (what you currently do). A much better and in my mind simpler way is to track the front AND the back of the relevant data within your deque.
As a simple example of the above concept consider a static (cannot grow) deque of size 10:
class Deque
{
public:
Deque()
: front(0)
, count(0) {}
private:
size_t front;
size_t count;
enum {
MAXSIZE = 10
};
int data[MAXSIZE];
};
You can of course implement this and allow it to grow in size etc. But for simplicity I'm leaving all that out. Now to allow a user to add to the deque:
void Deque::push_back(int value)
{
if(count>=MAXSIZE)
throw std::runtime_error("Deque full!");
data[(front+count)%MAXSIZE] = value;
count++;
}
And to pop off the back:
int Deque::pop_back()
{
if(count==0)
throw std::runtime_error("Deque empty! Cannot pop!");
int value = data[(front+(--count))%MAXSIZE];
return value;
}
Now the key thing to observe in the above functions is how we are accessing the data within the array. By modding with MAXSIZE we ensure that we are not accessing out of bounds, and that we are hitting the right value. Also as the value of front changes (due to push_front, pop_front) the modulus operator ensures that wrap around is dealt with appropriately. I'll show you how to do push_front, you can figure out pop_front for yourself:
void Deque::push_front(int value)
{
if(count>=MAXSIZE)
throw std::runtime_error("Deque full!");
// Determine where front should now be.
if (front==0)
front = MAXSIZE-1;
else
--front;
data[front] = value;
++count;
}