What STL algorithm can determine if exactly one item in a container satisfies a predicate?

Question

I need an STL algorithm that takes a predicate and a collection and returns true if one and only one member of the collection satisfies the predicate, otherwise returns false.

How would I do this using STL algorithms?

E.g., to replace the following with STL algorithm code to express the same return value.

int count = 0;

for( auto itr = c.begin(); itr != c.end(); ++itr ) {
    if ( predicate( *itr ) ) {
      if ( ++count > 1 ) {
        break;
      }
    }
}

return 1 == count;

Answer 1

Two things come to my mind:

std::count_if and then compare the result to 1.

To avoid traversing the whole container in case eg the first two elements already match the predicate I would use two calls looking for matching elements. Something along the line of

auto it = std::find_if(begin,end,predicate);
if (it == end) return false;
++it;
return std::none_of(it,end,predicate);

Or if you prefer it more compact:

auto it = std::find_if(begin,end,predicate); 
return (it != end) && std::none_of(std::next(it),end,predicate);

Credits goes to Remy Lebeau for compacting, Deduplicator for debracketing and Blastfurnance for realizing that we can also use none_of the std algorithms.

Answer 2

You can use std::count_if^† to count and return if it is one.

For example:

#include <iostream>
#include <algorithm> // std::count_if
#include <vector>    // std::vector
#include <ios>       // std::boolalpha

template<class Iterator, class UnaryPredicate>
constexpr bool is_count_one(Iterator begin, const Iterator end, UnaryPredicate pred)
{
    return std::count_if(begin, end, pred) == 1;
}

int main()
{
    std::vector<int> vec{ 2, 4, 3 };
    // true: if only one Odd element present in the container
    std::cout << std::boolalpha
              << is_count_one(vec.cbegin(), vec.cend(),
                  [](const int ele) constexpr noexcept -> bool { return ele & 1; });
    return 0;
}

---

^†Update: However, std::count_if counts entire element in the container, which is not good as the algorithm given in the question. The best approach using the standard algorithm collections has been mentioned in @formerlyknownas_463035818 's answer.

That being said, OP's approach is also good as the above mentioned best standard approach, where a short-circuiting happens when count reaches 2. If someone is interested in a non-standard algorithm template function for OP's approach, here is it.

#include <iostream>
#include <vector>    // std::vector
#include <ios>       // std::boolalpha
#include <iterator>  // std::iterator_traits

template<class Iterator, class UnaryPredicate>
bool is_count_one(Iterator begin, const Iterator end, UnaryPredicate pred)
{
    typename std::iterator_traits<Iterator>::difference_type count{ 0 };
    for (; begin != end; ++begin) {
        if (pred(*begin) && ++count > 1) return false;
    }
    return count == 1;
}

int main()
{
    std::vector<int> vec{ 2, 3, 4, 2 };
    // true: if only one Odd element present in the container
    std::cout << std::boolalpha
              << is_count_one(vec.cbegin(), vec.cend(),
                  [](const int ele) constexpr noexcept -> bool { return ele & 1; });
    return 0;
}

Now that can be generalized, by providing one more parameter, the number of N element(s) has/ have to be found in the container.

template<typename Iterator>
using diff_type = typename std::iterator_traits<Iterator>::difference_type;

template<class Iterator, class UnaryPredicate>
bool has_exactly_n(Iterator begin, const Iterator end, UnaryPredicate pred, diff_type<Iterator> N = 1)
{
    diff_type<Iterator> count{ 0 };
    for (; begin != end; ++begin) {
        if (pred(*begin) && ++count > N) return false;
    }
    return count == N;
}

Answer 3

Starting from formerlyknownas_463035818's answer, this can be generalized to seeing if a container has exactly n items that satisfy a predicate. Why? Because this is C++ and we're not satisfied until we can read email at compile time.

template<typename Iterator, typename Predicate>
bool has_exactly_n(Iterator begin, Iterator end, size_t count, Predicate predicate)
{
    if(count == 0)
    {
        return std::none_of(begin, end, predicate);
    }
    else
    {
        auto iter = std::find_if(begin, end, predicate);
        return (iter != end) && has_exactly_n(std::next(iter), end, count - 1, predicate);
    }
}

Answer 4

Using std::not_fn to negate a predicate

As the core of the algorithm of this question (as has been elegantly covered by combining std::find_if and std::none_of in the accepted answer), with short-circuiting upon failure, is to scan a container for a unary predicate and, when met, continue scanning the rest of the container for the negation of the predicate, I will mention also the negator std::not_fn introduced in C++17, replacing the less useful std::not1 and std::not2 constructs.

We may use std::not_fn to implement the same predicate logic as the accepted answer (std::find_if conditionally followed by std::none_of), but with somewhat different semantics, replacing the latter step (std::none_of) with std::all_of over the negation of the unary predicate used in the first step (std::find_if). E.g.:

// C++17
#include <algorithm>   // std::find_if
#include <functional>  // std::not_fn
#include <ios>         // std::boolalpha
#include <iostream>
#include <iterator>  // std::next
#include <vector>

template <class InputIt, class UnaryPredicate>
constexpr bool one_of(InputIt first, InputIt last, UnaryPredicate p) {
  auto it = std::find_if(first, last, p);
  return (it != last) && std::all_of(std::next(it), last, std::not_fn(p));
}

int main() {
  const std::vector<int> v{1, 3, 5, 6, 7};
  std::cout << std::boolalpha << "Exactly one even number : "
            << one_of(v.begin(), v.end(), [](const int n) {
                 return n % 2 == 0;
               });  // Exactly one even number : true
}

---

A parameter pack approach for static size containers

As I’ve already limited this answer to C++14 (and beyond), I’ll include an alternative approach for static size containers (here applied for std::array, specifically), making use of std::index_sequence combined with parameter pack expansion:

#include <array>
#include <ios>         // std::boolalpha
#include <iostream>
#include <utility>     // std::(make_)index_sequence

namespace detail {
template <typename Array, typename UnaryPredicate, std::size_t... I>
bool one_of_impl(const Array& arr, const UnaryPredicate& p,
                 std::index_sequence<I...>) {
  bool found = false;
  auto keep_searching = [&](const int n){
      const bool p_res = found != p(n);
      found = found || p_res;
      return !found || p_res;
  };
  return (keep_searching(arr[I]) && ...) && found;
}
}  // namespace detail

template <typename T, typename UnaryPredicate, std::size_t N,
          typename Indices = std::make_index_sequence<N>>
auto one_of(const std::array<T, N>& arr,
            const UnaryPredicate& p) {
  return detail::one_of_impl(arr, p, Indices{});
}

int main() {
  const std::array<int, 5> a{1, 3, 5, 6, 7};
  std::cout << std::boolalpha << "Exactly one even number : "
            << one_of(a, [](const int n) {
                 return n % 2 == 0;
               });  // Exactly one even number : true
}

This will also short-circuit upon early failure (“found more than one”), but will contain a few more simple boolean comparisons than in the approach above.

However, note that this approach could have its draw-backs, particularly for optimized code for container inputs with many elements, as is pointed out by @PeterCordes in a comment below. Citing the comment (as comments are not guaranteed to persist over time):

Just because the size is static doesn't mean that fully unrolling the loop with templates is a good idea. In the resulting asm, this needs a branch every iteration anyway to stop on found, so that might as well be a loop-branch. CPUs are good at running loops (code caches, loopback buffers). Compilers will fully unroll static-sized loops based on heuristics, but probably won't roll this back up if a is huge. So your first one_of implementation has the best of both worlds already, assuming a normal modern compiler like gcc or clang, or maybe MSVC