I thought the following would work but it just outputs zero. Ideas?
std::vector<int> a = { 1, 2, 3 };
std::vector<int> b = { 4, 5, 6 };
int max = *std::max(std::max(a.begin(), a.end()), std::max(b.begin(), b.end()));
std::cout << max;
You're using std::max, which compares its arguments. That is, it's returning the greater of the two iterators.
What you want for the inner invocation is std::max_element, which finds the maximum element in a range:
std::vector<int> a = { 1, 2, 3 };
std::vector<int> b = { 4, 5, 6 };
int max = std::max(*std::max_element(a.begin(), a.end()), *std::max_element(b.begin(), b.end()));
std::cout << max;
Live example
As #MikeSeymour correctly pointed out in comments, the above code assumes the ranges are not empty, as it unconditionally dereferences the iterators returned from std::max_element. If one of the ranges was empty, the returned iterator would be the past-the-end one, which cannot be dereferenced.
Here's a way that behaves sensibly with empty ranges. If either range is empty, you still get the maximum from the other range. If both ranges are empty, you get INT_MIN.
int m = std::accumulate(begin(b), end(b),
std::accumulate(begin(a), end(a), INT_MIN, std::max<int>),
std::max<int>);
std::accumulate is better here, since you want a value, not an iterator, as the result.
int m = std::max(std::max_element(a.begin(), a.end()), std::max_element(b.begin(), b.end()));
This finds maximum of the maximums of the individual vectors. For example, for 1st vector, { 1, 2, 3 }, max value is 3, and for 2nd vector, { 4, 5, 6 }, max value is 6, max of 3 and 6 is now 6
Related
Let's say I have a a vector<vector<int>>. I want to use ranges::transform in such a way that I get
vector<vector<int>> original_vectors;
using T = decltype(ranges::views::transform(original_vectors[0], [&](int x){
return x;
}));
vector<int> transformation_coeff;
vector<T> transformed_vectors;
for(int i=0;i<n;i++){
transformed_vectors.push_back(ranges::views::transform(original_vectors[i], [&](int x){
return x * transformation_coeff[i];
}));
}
Is such a transformation, or something similar, currently possible in C++?
I know its possible to simply store the transformation_coeff, but it's inconvenient to apply it at every step. (This will be repeated multiple times so it needs to be done in O(log n), therefore I can't explicitly apply the transformation).
Yes, you can have a vector of ranges. The problem in your code is that you are using a temporary lambda in your using statement. Because of that, the type of the item you are pushing into the vector later is different from T. You can solve it by assigning the lambda to a variable first:
vector<vector<int>> original_vectors;
auto lambda = [&](int x){return x;};
using T = decltype(ranges::views::transform(original_vectors[0], lambda));
vector<T> transformed_vectors;
transformed_vectors.push_back(ranges::views::transform(original_vectors[0], lambda));
It is not possible in general to store different ranges in a homogeneous collection like std::vector, because different ranges usually have different types, especially if transforms using lambdas are involved. No two lambdas have the same type and the type of the lambda will be part of the range type. If the signatures of the functions you want to pass to the transform are the same, you could wrap the lambdas in std::function as suggested by #IlCapitano (https://godbolt.org/z/zGETzG4xW). Note that this comes at the cost of the additional overhead std::function entails.
A better option might be to create a range of ranges.
If I understand you correctly, you have a vector of n vectors, e.g.
std::vector<std::vector<int>> original_vector = {
{1, 5, 10},
{2, 4, 8},
{5, 10, 15}
};
and a vector of n coefficients, e.g.
std::vector<int> transformation_coeff = {2, 1, 3};
and you want a range of ranges representing the transformed vectors, where the ith range represents the ith vector's elements which have been multiplied by the ith coefficient:
{
{ 2, 10, 20}, // {1, 5, 10} * 2
{ 2, 4, 8}, // {2, 4, 8} * 1
{15, 30, 45} // {5, 10, 15} * 3
}
Did I understand you correctly? If yes, I don't understand what you mean with your complexity requirement of O(log n). What does n refer to in this scenario? How would this calculation be possible in less than n steps? Here is a solution that gives you the range of ranges you want. Evaluating this range requires O(n*m) multiplications, where m is an upper bound for the number of elements in each inner vector. I don't think it can be done in less steps because you have to multiply each element in original_vector once. Of course, you can always just evaluate part of the range, because the evaluation is lazy.
C++20
The strategy is to first create a range for the transformed i-th vector given the index i. Then you can create a range of ints using std::views::iota and transform it to the inner ranges:
auto transformed_ranges = std::views::iota(0) | std::views::transform(
[=](int i){
// get a range containing only the ith inner range
auto ith = original_vector | std::views::drop(i) | std::views::take(1) | std::views::join;
// transform the ith inner range
return ith | std::views::transform(
[=](auto const& x){
return x * transformation_coeff[i];
}
);
}
);
You can now do
for (auto const& transformed_range : transformed_ranges){
for (auto const& val : transformed_range){
std::cout << val << " ";
}
std::cout<<"\n";
}
Output:
2 10 20
2 4 8
15 30 45
Full Code on Godbolt Compiler Explorer
C++23
This is the perfect job for C++23's std::views::zip_transform:
auto transformed_ranges = std::views::zip_transform(
[=](auto const& ith, auto const& coeff){
return ith | std::views::transform(
[=](auto const& x){
return x * coeff;
}
);
},
original_vector,
transformation_coeff
);
It's a bit shorter and has the added benefit that transformation_coeff is treated as a range as well:
It is more general, because we are not restricted to std::vectors
In the C++20 solution you get undefined behaviour without additional size checking if transformation_coeff.size() < original_vector.size() because we are indexing into the vector, while the C++23 solution would just return a range with fewer elements.
Full Code on Godbold Compiler Explorer
For example:
Let us have a sorted vector with elements: [1, 3, 4, 6, 7, 10, 11, 13]
And we have an element N = 5
I want output as:
4
Since 4 is the greatest element smaller than N.
I want to modify Binary Search to get the answer
What would you want to happen if there is an element that equals N in the vector?
I would use std::lower_bound (or std::upper_bound depending on the answer to the above question). It runs in logarithmic time which means it's probably using binary search under the hood.
std::optional<int> find_first_less_than(int n, std::vector<int> data) {
// things must be sorted before processing
std::sort(data.begin(), data.end());
auto it = std::lower_bound(data.begin(), data.end(), n);
// if all of the elements are above N, we'll return nullopt
if (it == data.begin()) return std::nullopt;
return *std::prev(it);
}
I have a complex problem and have been trying to identify what needs to be a very, very efficient algorithm. I'm hoping i can get some ideas from you helpful folks. Here is the situation.
I have a vector of vectors. These nested vectors are of various length, all storing integers in a random order, such as (pseudocode):
vector_list = {
{ 1, 4, 2, 3 },
{ 5, 9, 2, 1, 3, 3 },
{ 2, 4, 2 },
...,
100 more,
{ 8, 2, 2, 4 }
}
and so on, up to over 100 different vectors at a time inside vector_list. Note that the same integer can appear in each vector more than once. I need to remove from this vector_list any vectors that are duplicates of another vector. A vector is a duplicate of another vector if:
It has the same integers as the other vector (regardless of order). So if we have
vec1 = { 1, 2, 3 }
vec2 = { 2, 3, 1 }
These are duplicates and I need to remove one of them, it doesnt matter which one.
A vector contains all of the other integers of the other vector. So if we have
vec1 = { 3, 2, 2 }
vec2 = { 4, 2, 3, 2, 5 }
Vec2 has all of the ints of vec1 and is bigger, so i need to delete vec1 in favor of vec2
The problem is as I mentioned the list of vectors can be very big, over 100, and the algorithm may need to run as many as 1000 times on a button click, with a different group of 100+ vectors over 1000 times. Hence the need for efficiency. I have considered the following:
Sorting the vectors may make life easier, but as I said, this has to be efficient, and i'd rather not sort if i didnt have to.
It's more complicated by the fact that the vectors aren't in any order with respect to their size. For example, if the vectors in the list were ordered by size:
vector_list = {
{ },
{ },
{ },
{ },
{ },
...
{ },
{ }
}
It might make life easier, but that seems like it would take a lot of effort and I'm not sure about the gain.
The best effort I've had so far to try and solve this problem is:
// list of vectors, just 4 for illustration, but in reality more like 100, with lengths from 5 to 15 integers long
std::vector<std::vector<int>> vector_list;
vector_list.push_back({9});
vector_list.push_back({3, 4, 2, 8, 1});
vector_list.push_back({4, 2});
vector_list.push_back({1, 3, 2, 4});
std::vector<int>::iterator it;
int i;
int j;
int k;
// to test if a smaller vector is a duplicate of a larger vector, i copy the smaller vector, then
// loop through ints in the larger vector, seeing if i can find them in the copy of the smaller. if i can,
// i remove the item from the smaller copy, and if the size of the smaller copy reaches 0, then the smaller vector
// was a duplicate of the larger vector and can be removed.
std::vector<int> copy;
// flag for breaking a for loop below
bool erased_i;
// loop through vector list
for ( i = 0; i < vector_list.size(); i++ )
{
// loop again, so we can compare every vector to every other vector
for ( j = 0; j < vector_list.size(); j++ )
{
// don't want to compare a vector to itself
if ( i != j )
{
// if the vector in i loop is at least as big as the vector in j loop
if ( vector_list[i].size() >= vector_list[j].size() )
{
// copy the smaller j vector
copy = vector_list[j];
// loop through each item in the larger i vector
for ( k = 0; k < vector_list[i].size(); k++ ) {
// if the item in the larger i vector is in the smaller vector,
// remove it from the smaller vector
it = std::find(copy.begin(), copy.end(), vector_list[i][k]);
if (it != copy.end())
{
// erase
copy.erase(it);
// if the smaller vector has reached size 0, then it must have been a smaller duplicate that
// we can delete
if ( copy.size() == 0 ) {
vector_list.erase(vector_list.begin() + j);
j--;
}
}
}
}
else
{
// otherwise vector j must be bigger than vector i, so we do the same thing
// in reverse, trying to erase vector i
copy = vector_list[i];
erased_i = false;
for ( k = 0; k < vector_list[j].size(); k++ ) {
it = std::find(copy.begin(), copy.end(), vector_list[j][k]);
if (it != copy.end()) {
copy.erase(it);
if ( copy.size() == 0 ) {
vector_list.erase(vector_list.begin() + i);
// put an extra flag so we break out of the j loop as well as the k loop
erased_i = true;
break;
}
}
}
if ( erased_i ) {
// break the j loop because we have to start over with whatever
// vector is now in position i
break;
}
}
}
}
}
std::cout << "ENDING VECTORS\n";
// TERMINAL OUTPUT:
vector_list[0]
[9]
vector_list[1]
[3, 4, 2, 8, 1]
So this function gives me the right results, as these are the 2 unique vectors. It also gives me the correct results if i push the initial 4 vectors in reverse order, so the smallest one comes last for example. But it feels so inefficient comparing every vector to every other vector. Plus i have to create these "copies" and try to reduce them to 0 .size() with every comparison I make. very inefficient.
Anyways, any ideas on how I could make this speedier would be much appreciated. Maybe some kind of organization by vector length, I dunno.... It seems wasteful to compare them all to each other.
Thanks!
Loop through the vectors and for each vector, map the count of unique values occurring in it. unordered_map<int, int> would suffice for this, let's call it M.
Also maintain a set<unordered_map<int, int>>, say S, ordered by the size of unordered_map<int, int> in decreasing order.
Now we will have to compare contents of M with the contents of unordered_maps in S. Let's call M', the current unordered_map in S being compared with M. M will be a subset of M' only when the count of all the elements in M is less than or equal to the count of their respective elements in M'. If that's the case then it's a duplicate and we'll not insert. For any other case, we'll insert. Also notice that if the size of M is greater than the size of M', M can't be a subset of M'. That means we can insert M in S. This can be used as a pre-condition to speed things up. Maintain the indices of vectors which weren't inserted in S, these are the duplicates and have to be deleted from vector_list in the end.
Time Complexity: O(N*M) + O(N^2*D) + O(N*log(N)) = O(N^2*D) where N is the number of vectors in vector_list, M is the average size of the vectors in vector_list and D is the average size of unordered_map's in S. This is for the worst case when there aren't any duplicates. For average case, when there are duplicates, the second complexity will come down.
Edit: The above procedure will create a problem. To fix that, we'll need to make unordered_maps of all vectors, store them in a vector V, and sort that vector in decreasing order of the size of unordered_map. Then, we'll start from the biggest in this vector and apply the above procedure on it. This is necessary because, a subset, say M1 of a set M2, can be inserted into S before M2 if the respective vector of M1 comes before the respective vector of M2 in vector_list. So now we don't really need S, we can compare them within V itself. Complexity won't change.
Edit 2: The same problem will occur again if sizes of two unordered_maps are the same in V when sorting V. To fix that, we'll need to keep the contents of unordered_maps in some order too. So just replace unordered_map with map and in the comparator function, if the size of two maps is the same, compare element by element and whenever the keys are not the same for the very first time or are same but the M[key] is not the same, put the bigger element before the other in V.
Edit 3: New Time Complexity: O(N*M*log(D)) + O(N*D*log(N)) + O(N^2*D*log(D)) = O(N^2*D*log(D)). Also you might want to pair the maps with the index of the respective vectors in vector_list so as to know which vector you must delete from vector_list when you find a duplicate in V.
IMPORTANT: In sorted V, we must start checking from the end just to be safe (in case we choose to delete a duplicate from vector_list as well as V whenever we encounter it). So for the last map in V compare it with the rest of the maps before it to check if it is a duplicate.
Example:
vector_list = {
{1, 2, 3},
{2, 3, 1},
{3, 2, 2},
{4, 2, 3, 2, 5},
{1, 2, 3, 4, 6, 2},
{2, 3, 4, 5, 6},
{1, 5}
}
Creating maps of respective vectors:
V = {
{1->1, 2->1, 3->1},
{1->1, 2->1, 3->1},
{2->2, 3->1},
{2->2, 3->1, 4->1, 5->1},
{1->1, 2->2, 3->1, 4->1, 6->1},
{2->1, 3->1, 4->1, 5->1, 6->1},
{1->1, 5->1}
}
After sorting:
V = {
{1->1, 2->2, 3->1, 4->1, 6->1},
{2->1, 3->1, 4->1, 5->1, 6->1},
{2->2, 3->1, 4->1, 5->1},
{1->1, 2->1, 3->1},
{1->1, 2->1, 3->1},
{1->1, 5->1},
{2->2, 3->1}
}
After deleting duplicates:
V = {
{1->1, 2->2, 3->1, 4->1, 6->1},
{2->1, 3->1, 4->1, 5->1, 6->1},
{2->2, 3->1, 4->1, 5->1},
{1->1, 5->1}
}
Edit 4: I tried coding it up. Running it a 1000 times on a list of 100 vectors, the size of each vector being in range [1-250], the range of the elements of vector being [0-50] and assuming the input is available for all the 1000 times, it takes around 2 minutes on my machine. It goes without saying that there is room for improvement in my code (and my machine).
My approach is to copy the vectors that pass the test to an empty vector.
May be inefficient.
May have bugs.
HTH :)
C++ Fiddle
#include <algorithm>
#include <iostream>
#include <iterator>
#include <vector>
int main(int, char **) {
using namespace std;
using vector_of_integers = vector<int>;
using vector_of_vectors = vector<vector_of_integers>;
vector_of_vectors in = {
{ 1, 4, 2, 3 }, // unique
{ 5, 9, 2, 1, 3, 3 }, // unique
{ 3, 2, 1 }, // exists
{ 2, 4, 2 }, // exists
{ 8, 2, 2, 4 }, // unique
{ 1, 1, 1 }, // exists
{ 1, 2, 2 }, // exists
{ 5, 8, 2 }, // unique
};
vector_of_vectors out;
// doesnt_contain_vector returns true when there is no entry in out that is superset of any of the passed vectors
auto doesnt_contain_vector = [&out](const vector_of_integers &in_vector) {
// is_subset returns true a vector contains all the integers of the passed vector
auto is_subset = [&in_vector](const vector_of_integers &out_vector) {
// contained returns true when the vector contains the passed integer
auto contained = [&out_vector](int i) {
return find(out_vector.cbegin(), out_vector.cend(), i) != out_vector.cend();
};
return all_of(in_vector.cbegin(), in_vector.cend(), contained);
};
return find_if(out.cbegin(), out.cend(), is_subset) == out.cend();
};
copy_if(in.cbegin(), in.cend(), back_insert_iterator<vector_of_vectors>(out), doesnt_contain_vector);
// show results
for (auto &vi: out) {
copy(vi.cbegin(), vi.cend(), std::ostream_iterator<int>(std::cout, ", "));
cout << "\n";
}
}
You could try something like this. I use std::sort and std::includes. Perhaps this is not the most effective solution.
// sort all nested vectors
std::for_each(vlist.begin(), vlist.end(), [](auto& v)
{
std::sort(v.begin(), v.end());
});
// sort vector of vectors by length of items
std::sort(vlist.begin(), vlist.end(), [](const vector<int>& a, const vector<int>& b)
{
return a.size() < b.size();
});
// exclude all duplicates
auto i = std::begin(vlist);
while (i != std::end(vlist)) {
if (any_of(i+1, std::end(vlist), [&](const vector<int>& a){
return std::includes(std::begin(a), std::end(a), std::begin(*i), std::end(*i));
}))
i = vlist.erase(i);
else
++i;
}
I want to check whether given two integers in a specific order are in the same order in a given integer array.
I wonder whether there is an easy way to do this like a built-in CPP method.
If there is no built-in method, suggest me an efficient way to do this as I have a few sets of two integers (not only one set) to check over one array.
given two numbers: 8 3
given array: 2 8 6 1 3 9
output: YES
You could do something like
bool check(std::pair<int, int> numbers = {8, 3},
std::array<int, 6> arr = {2, 8, 6, 1, 3, 9}) {
if (numbers.first != numbers.second)
return std::find(std::find(std::begin(arr), std::end(arr), numbers.first), std::end(arr), numbers.second) == std::end(arr);
return std::count(std::begin(arr), std::end(arr), numbers.first) >= 2;
}
If both numbers are different the inner find searches for the first value. The outer find starts at the position of the first value and searches for the second value.
Else the count is checked.
You could also try:
std::array<int, 6> content = {2, 3, 6, 1, 8, 9};
auto lookup = [content](int a, int b)
{
return std::distance(std::find(content.begin(), content.end(), a), std::find(content.rbegin(), content.rend(), b));
};
lookup(8, 3);
lookup will be positive if 8 comes before 3 and negative otherwise.
Search the entire container to find the first one. Search from the position of the first one to the end of the container to find the second one. If that search succeeds, they're in the expected order. If not, they're not.
int first_value = 8;
int second_value = 3;
std::array<int, 6> values = { 2, 8, 6, 1, 3, 9 };
auto first_pos = std::find(values.begin(), values.end(), first_value);
if (first_pos != values.end())
++first_pos;
auto second_pos = std::find(first_pos, values.end(), second_value);
if (second_pos != values.end())
std::cout << "YES\n";
Use adjacent find. I suppose find_if could also do the job.
A question that might appear trivial, but I am wondering if there's a way of obtaining the count of integers made unique after I transform an array containing repeated integers into an unordered_set. To be clear, I start with some array, turn into an unordered set, and suddenly, the unordered_set only contains unique integers, and I am simply after the repeat number of the integers in the unordered_set.
Is this possible at all? (something like unordered_set.count(index) ?)
A question that might appear trivial, but I am wondering if there's a way of obtaining the count of integers made unique after I transform an array containing repeated integers into an unordered_set.
If the container is contiguous, like an array, then I believe you can use ptrdiff_t to count them after doing some iterator math. I'm not sure about non-contiguous containers, though.
Since you start with an array:
Call unique on the array
unique returns iter.end()
Calculate ptrdiff_t count using iter.begin() and iter.end()
Remember that the calculation in step 3 needs to be adjusted for the sizeof and element.
But to paraphrase Beta, some containers lend themselves to this, and other do not. If you have an unordered set (or a map or a tree), then the information will not be readily available.
According to your answer to the user2357112's question I will write a solution.
So, let's assume that instead of unordered_set we will use a vector and our vector has values like this:
{1, 1, 1, 3, 4, 1, 1, 4, 4, 5, 5};
So, we want to get numbers (in different vector I think) of how many times particular value appears in the vector, right? And in this specific case result would be: 1 appears 5 times, 3 appears one time, 4 appears 3 times and 5 appears 2 times.
To get this done, one possible solution can be like this:
Get unique entries from source vector and store them in different vector, so this vector will contain: 1, 3, 4, 5
Iterate through whole unique vector and count these elements from source vector.
Print result
The code from point 1, can be like this:
template <typename Type>
vector<Type> unique_entries (vector<Type> vec) {
for (auto iter = vec.begin (); iter != vec.end (); ++iter) {
auto f = find_if (iter+1, vec.end (), [&] (const Type& val) {
return *iter == val;
});
if (f != vec.end ()) {
vec.erase (remove (iter+1, vec.end (), *iter), vec.end ());
}
}
return vec;
}
The code from point 2, can be like this:
template <typename Type>
struct Properties {
Type key;
long int count;
};
template <typename Type>
vector<Properties<Type>> get_properties (const vector<Type>& vec) {
vector<Properties<Type>> ret {};
auto unique_vec = unique_entries (vec);
for (const auto& uv : unique_vec) {
auto c = count (vec.begin (), vec.end (), uv); // (X)
ret.push_back ({uv, c});
}
return ret;
}
Of course we do not need Properties class to store key and count value, you can return just a vector of int (with count of elements), but as I said, it is one of the possible solutions. So, by using unique_entries we get a vector with unique entries ( :) ), then we can iterate through the whole vector vec (get_properties, using std::count marked as (X)), and push_back Properties object to the vector ret.
The code from point 3, can be like this:
template <typename Type>
void show (const vector<Properties<Type>>& vec) {
for (const auto& v : vec) {
cout << v.key << " " << v.count << endl;
}
}
// usage below
vector<int> vec {1, 1, 1, 3, 4, 1, 1, 4, 4, 5, 5};
auto properties = get_properties (vec);
show (properties);
And the result looks like this:
1 5
3 1
4 3
5 2
What is worth to note, this example has been written using templates to provide flexibility of choosing type of elements in the vector. If you want to store values of long, long long, short, etc, instead of int type, all you have to do is to change definition of source vector, for example:
vector<unsigned long long> vec2 {1, 3, 2, 3, 4, 4, 4, 4, 3, 3, 2, 3, 1, 7, 2, 2, 2, 1, 6, 5};
show (get_properties (vec2));
will produce:
1 3
3 5
2 5
4 4
7 1
6 1
5 1
which is desired result.
One more note, you can do this with vector of string as well.
vector<string> vec_str {"Thomas", "Rick", "Martin", "Martin", "Carol", "Thomas", "Martin", "Josh", "Jacob", "Jacob", "Rick"};
show (get_properties (vec_str));
And result is:
Thomas 2
Rick 2
Martin 3
Carol 1
Josh 1
Jacob 2
I assume you're trying to get a list of unique values AND the number of their occurences. If that's the case, then std::map provides the cleanest and simplest solution:
//Always prefer std::vector (or at least std::array) over raw arrays if you can
std::vector<int> myInts {2,2,7,8,3,7,2,3,46,7,2,1};
std::map<int, unsigned> uniqueValues;
//Get unique values and their count
for (int val : myInts)
++uniqueValues[val];
//Output:
for (const auto & val : uniqueValues)
std::cout << val.first << " occurs " << val.second << " times." << std::endl;
In this case it doesn't have to be std::unordered_set.