function for finding last item less-th

Question

Is there a function in that uses binary search, like lower_bound but that returns the last item less-than-or-equal-to according to a given predicate?

lower_bound is defined to:

Finds the position of the first element in an ordered range that has a value greater than or equivalent to a specified value, where the ordering criterion may be specified by a binary predicate.

and upper_bound:

Finds the position of the first element in an ordered range that has a value that is greater than a specified value, where the ordering criterion may be specified by a binary predicate.

Specifically, I have a container of time ordered events and for a given time I want to find the last item that came before or at that point. Can I achieve this with some combination of upper/lower bound, reverse iterators and using std::greater or std::greater_equal ?

EDIT: A tweak was needed to user763305's suggestion to cope with if you ask for a point before the start of the array:

iterator it=upper_bound(begin(), end(), val, LessThanFunction());
if (it!=begin()) {
  it--; // not at end of array so rewind to previous item
} else {
  it=end(); // no items before this point, so return end()
}
return it;

Answer 1

Here is a wrapper function around upper_bound which returns the largest number in a container or array which is less than or equal to a given value:

template <class ForwardIterator, class T>
  ForwardIterator largest_less_than_or_equal_to ( ForwardIterator first, 
                                                  ForwardIterator last,
                                                  const T& value)
{
  ForwardIterator upperb = upper_bound(first, last, value);

  // First element is >, so none are <=
  if(upperb == first)
    return NULL;

  // All elements are <=, so return the largest.
  if(upperb == last)
    return --upperb;

  return upperb - 1;
}

For a better explanation of what this is doing and how to use this function, check out:

C++ STL — Find last number less than or equal to a given element in an array or container

Answer 2

I have test your reverse iterator solution, it is correct.

Given v is sorted by '<'

Find last element less than x:

auto iter = std::upper_bound(v.rbegin(), v.rend(), x, std::greater<int>());
if(iter == v.rend())
    std::cout<<"no found";
else
    std::cout<<*iter;

Find last element less than equal x:

auto iter = std::lower_bound(v.rbegin(), v.rend(), x, std::greater<int>());
if(iter == v.rend())
    std::cout<<"no found";
else
    std::cout<<*iter;

This is better than iter -= 1 version

Answer 3

In a sorted container, the last element that is less than or equivalent to x, is the element before the first element that is greater than x.

Thus you can call std::upper_bound, and decrement the returned iterator once. (Before decrementing, you must of course check that it is not the begin iterator; if it is, then there are no elements that are less than or equivalent to x.)