I've got a set and I want to find the largest number not greater than x in it. (something like lower_bound(x) ) how should i do it? Is there any predefined functions?
set<int> myset;
myset.insert(blahblahblah);
int y;
//I want y to be greatest number in myset not greater than x
You can use upper_bound like this: upper_bound(x)--. Upper bound gives you the first element greater than x, so the element you seek is the one before that. You need a special case if upper_bound returns begin().
In addition to lower_bound there is also upper_bound
C++ reference
The function returns an iterator to the first value that is strictly greater than yours. If it returns begin() then all of them are, otherwise subtract one from the resulting iterator to get the value you are looking for.
Related
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How do I get the index of an iterator of an std::vector?
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I found some c++ code that I would like to understand. In this code they use
int airplane = min_element(min_cost_airplane.begin(),
min_cost_airplane.end()) - min_cost_airplane.begin();
But I don't know what this line of code exactly accomplishes. min_cost_airplane is a vector. I understand the min_element function, but I can't wrap my head around the -vector.begin at the end. Is the structure of this line of code common used? The thing I understand is that this line of code returns an iterator to the smallest element in the vector minus an iterator to the first element of the vector. So what does the iterator point to?
Can someone please help me?
The std::min_element algorithm returns an iterator. You can dereference that iterator to access the "minimum" element of the container. If, instead, you want to know the index of the element you need to compute it as the distance from the beginning of the container.
For random-access iterators you can subtract the iterators to get the offset, or index value. That's what your example does. There's also a std::distance function that computes the index but it also works for non-random access iterators. For example:
auto iter = std::min_element(min_cost_airplane.begin(), min_cost_airplane.end());
int index = std::distance(min_cost_airplane.begin(), iter);
std::min_element returns an iterator to the first instance of the minimum value in the given range.
begin returns an iterator to the first element in the container.
The std::vector iterators are special (it's a random access iterator) in that you can subtract one from another which yields the distance between them in terms of elements (it's pointer arithmetic under the hood). To be more generic and clearer, write
auto airplane = std::distance(
min_cost_airplane.begin(),
std::min_element(min_cost_airplane.begin(), min_cost_airplane.end())
);
std::min_element is used to finds the smallest element in the range [first, last].
std::vector<int> v{3, 8, 4, 2, 5, 9};
std::vector<int>::iterator result = std::min_element(v.begin(), v.end());
//result iterator to the minimum value in vector v.
std::cout << "min element is: " << *result;
output: min element is: 2
Note : The smallest value in vector v is 2.
I am trying to determine if emplace_hint should be used to insert a key into a multimap (as opposed to regular emplace). I have already calculated the range of the key in an earlier operation (on the same key):
range = multimap.equal_range(key);
Should I use range.first, range.second, or nothing as a hint to insert the key, value pair? What if the range is empty?
Should I use range.first, range.second, or nothing as a hint to insert the key, value pair?
As std::multimap::emplace_hint() states:
Inserts a new element into the container as close as possible to the position just before hint.
(emphasis is mine) you should use second iterator from range and it should make insertion more efficient:
Complexity
Logarithmic in the size of the container in general, but amortized constant if the new element is inserted just before hint.
as for empty range, it is still fine to use second iterator as it should always point to greater than element or behind the last if not such one exists.
First, performance wise, it will not make any difference if you use range.first or range.second. Let's have a look at the return value of equal_range:
std::equal_range - return value
std::pair containing a pair of iterators defining the wanted range,
the first pointing to the first element that is not less than value
and the second pointing to the first element greater than value. If
there are no elements not less than value, last is returned as the
first element. Similarly if there are no elements greater than value,
last is returned as the second element
This means that - when obtained for a value key - both range.first and range.secod are represent positions wherekeymay be correctly inserted right before. So performance wise it should not matter if you userange.firstorrange.last`. Complexity should be "amortized constant", since the new element is inserted just before hint.
Second, when the range is "empty", range.first and range.second are both one-past-the-end, and therefore performance as well as result are identical, actually the same as if you used emplace without any hint.
See the following program demonstrating this:
int main()
{
std::multimap<std::string, std::string> m;
// some clutter:
m.emplace(std::make_pair(std::string("k"), std::string("1")));
m.emplace(std::make_pair(std::string("k"), std::string("2")));
m.emplace(std::make_pair(std::string("z"), std::string("1")));
m.emplace(std::make_pair(std::string("z"), std::string("2")));
// relevant portion of demo data: order a-c-b may be preserved
m.emplace(std::make_pair(std::string("x"), std::string("a")));
m.emplace(std::make_pair(std::string("x"), std::string("c")));
m.emplace(std::make_pair(std::string("x"), std::string("b")));
auto r = m.equal_range("x");
// will insert "x.zzzz" before "x.a":
m.emplace_hint(r.first, std::make_pair(std::string("x"), std::string("zzzz")));
// will insert "x.0" right after "x.b":
m.emplace_hint(r.second, std::make_pair(std::string("x"), std::string("0")));
auto rEmpty = m.equal_range("e");
// "empty" range, normal lookup:
m.emplace_hint(rEmpty.first, std::make_pair(std::string("e"), std::string("b")));
m.emplace_hint(rEmpty.second, std::make_pair(std::string("e"), std::string("a")));
auto rWrong = m.equal_range("k");
m.emplace_hint(rWrong.first, std::make_pair(std::string("z"), std::string("a")));
for (const auto &p : m) {
std::cout << p.first << " => " << p.second << '\n';
}
}
Output:
e => b
e => a
k => 1
k => 2
x => zzzz
x => a
x => c
x => b
x => 0
z => a
z => 1
z => 2
In short: if you have a valid range for key pre-calculated, then use it when inserting key. It will help anyway.
EDIT:
There have been discussions around whether an "invalid" hint might lead to an insertion at a position that does not then reflect the "order of insertion" for values with the same key. This might be concluded from a general multimap statement "The order of the key-value pairs whose keys compare equivalent is the order of insertion and does not change. (since C++11)".
I did not find support for the one or the other point of view in any normative document. I just found the following statement in cplusplus multimap/emplace_hint documentation:
emplate <class... Args>
iterator emplace_hint (const_iterator position, Args&&... args);
position Hint for the position where the element can be inserted. The function optimizes its insertion time if position points to the
element that will follow the inserted element (or to the end, if it
would be the last). Notice that this does not force the new element to
be in that position within the multimap container (the elements in a
multimap always follow a specific order). const_iterator is a member
type, defined as a bidirectional iterator type that points to
elements.
I know that this is not a normative reference, but at least my Apple LLVM 8.0 compiler adheres to this definition (see demo above):
If one inserts an element with a "wrong" hint, i.e. one pointing even before the position where a pair shall be inserted, the algorithm recognizes this and chooses a valid position (see inserting "z"=>"a" where a hint points to an "x"-element).
If we use a range for key "x" and use range.first, the position right before the first x is interpreted as a valid position.
So: I think that m.emplace_hint(r.first,... behaves in a way that the algorithm chooses a valid position immediately, and that to a position close to hint overrules the general statement "The order of the key-value pairs whose keys compare equivalent is the order of insertion and does not change. (since C++11)".
I am relative new at C++ and I have little problem. I have vector and in that vector are vectors with 3 integers.
Inner vector represents like one person. 3 integers inside that inner vector represents distance from start, velocity and original index (because in input integers aren't sorted and in output I need to print original index not index in this sorted vector).
Now I have given some points representing distance from start and I need to find which person will be first at that point so I have been thinking that my first step would be that I would find closest person to the given point so basically I need to find lower_bound/upper_bound.
How can I use lower_bound if I want to find the lower_bound of first item in inner vectors? Or should I use struct/class instead of inner vectors?
You would use the version of std::lower_bound which takes a custom comparator (the versions marked "(2)" at the link); and you would write a comparator of vectors which compares vectors by their first item (or whatever other way you like).
Howerver:
As #doctorlove points out, std::lower_bound doesn't compare the vectors to each other, it compares them to a given value (be it a vector or a scalar). So it's possible you actually want to do something else.
It's usually not a good idea to keep fixed-length sequences of elements in std::vector's. Have you considered std::array?
It's very likely that your "vectors with 3 integers" actually stand for something else, e.g. points in a 3-dimensional geometric space; in which case, yes, they should be in some sort of class.
I am not sure that your inner things should be std::vector-s of 3 elements.
I believe that they should std::array-s of 3 elements (because you know that the size is 3 and won't change).
So you probably want to have
typedef std::array<double,3> element_ty;
then use std::vector<element_ty> and for the rest (your lower_bound point) do like in einpoklum's answer.
BTW, you probably want to use std::min_element with an explicit compare.
Maybe you want something like:
std::vector<element_ty> vec;
auto minit =
std::min_element(vec.begin(), vec.end(),
[](const element_ty& x, const element_ty&y) {
return x[0] < y[0]));
I'm looking for an STL sort that returns the element "closest" to the target value if the exact value is not present in the container. It needs to be fast, so essentially I'm looking for a slightly modified binary search... I could write it, but it seems like something that should already exist...
Do you mean the lower_bound/upper_bound functions? These perform a binary search and return the closest element above the value you're looking for.
Clarification: The global versions of lower/upper_bound only work if the range is sorted, as they use some kind of binary search internally. (Obviously, the lower/upper_bound methods in std::map always work). You said in your question that you were looking for some kind of binary search, so I'll assume the range is sorted.
Also, Neither lower_bound nor upper_bound returns the closest member. If the value X you're looking for isn't a member of the range, they will both return the first element greater then X. Otherwise, lower_bound will return the first value equal to X, upper_boundwill return the last value equals X.
So to find the closest value, you'd have to
call lower_bound
if it returns the end of the range, all values are less then X. The last (i.e. the highest) element is the closest one
it if returns the beginning of the range, all values are greater then X. The first (i.e. the lowest) element is the closest one
if it returns an element in the middle of the range, check that element and the element before - the one that's closer to X is the one you're looking for
So you're looking for an element which has a minimal distance from some value k?
Use std::transform to transform each x to x-k. The use std::min_element with a comparison function which returns abs(l) < abs(r). Then add k back onto the result.
EDIT: Alternatively, you could just use std::min_element with a comparison function abs(l-k) < abs(r-k), and eliminate the std::transform.
EDIT2: This is good for unsorted containers. For sorted containers, you probably want nikie's answer.
If the container is already sorted (as implied) you should be able to use std::upper_bound and the item directly before to figure out which is closest:
// Untested.
template <class Iter, class T>
Iter closest_value(Iter begin, Iter end, T value)
{
Iter result = std::upper_bound(begin, end, value);
if(result != begin)
{
Iter lower_result = result;
--lower_result;
if(result == end || ((value - *lower_result) < (*result - value)))
{
result = lower_result;
}
}
return result;
}
If the container is not sorted, use min_element with a predicate as already suggested.
If your data is not sorted, use std::min_element with a comparison functor that calculates your distance.
I have a std::set<int>, what's the proper way to find the largest int in this set?
What comparator are you using?
For the default this will work:
if(!myset.empty())
*myset.rbegin();
else
//the set is empty
This will also be constant time instead of linear like the max_element solution.
Sets are always ordered. Assuming you are using the default comparison (less), just grab the last element in the set. rbegin() might be useful.
I believe you are looking for std::max_element:
The max_element() function returns an
iterator to the largest element in the
range [start,end).
Since set sorts the element in ascending order by default, just pick up the last element in the set.
if(!myset.empty())
*myset.rend();
else
//the set is empty
In an ordered integer set, the last element is the largest one.
Before you push() in your set<int> save the value in int max in global variable