function for finding last item less-than-or-equal to, like lower_bound

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 predic

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

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.)

Answer 3

std::prev: https://en.cppreference.com/w/cpp/iterator/prev

#include <iostream>
#include <map>

int main()
{
    std::map<int, char> m{{2, 'a'}, {4, 'b'}, {6, 'c'}, {8, 'd'}, {10, 'e'}};

    int num = 3;
    auto it = m.upper_bound(num);
    auto pv = std::prev(it);

    std::cout << "upper bound of " << num << ": "
        << it->first << ", " << it->second << '\n';
    std::cout << "lower than or equal of " << num << ": "
        << pv->first << ", " << pv->second << '\n';
}

Output:

upper bound of 3: 4, b
lower than or equal than 3: 2, a

Answer 4

A simple implementation in Python:

def bi_search(arr, target):
    """
    index of the last element which <= target
    """

    lo, hi = 0, len(arr)
    while lo < hi:
        mid = lo + (hi - lo) // 2
        if arr[mid] <= target:
            lo = mid + 1
        else:
            hi = mid
    return lo - 1

Answer 5

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