set_intersection


Category: algorithms	Component type: function

Prototype

Set_intersection is an overloaded name; there are actually two set_intersection functions.

template <class InputIterator1, class InputIterator2, class OutputIterator>
OutputIterator set_intersection(InputIterator1 first1, InputIterator1 last1,
                                InputIterator2 first2, InputIterator2 last2,
                                OutputIterator result);

template <class InputIterator1, class InputIterator2, class OutputIterator,
          class StrictWeakOrdering>
OutputIterator set_intersection(InputIterator1 first1, InputIterator1 last1,
                                InputIterator2 first2, InputIterator2 last2,
                                OutputIterator result, 
                                StrictWeakOrdering comp);

Description

Set_intersection constructs a sorted range that is the intersection of the sorted ranges [first1, last1) and [first2, last2). The return value is the end of the output range.

In the simplest case, set_intersection performs the "intersection" operation from set theory: the output range contains a copy of every element that is contained in both [first1, last1) and [first2, last2). The general case is more complicated, because the input ranges may contain duplicate elements. The generalization is that if a value appears m times in [first1, last1) and n times in [first2, last2) (where m or n may be zero), then it appears min(m,n) times in the output range. [1] Set_intersection is stable, meaning both that elements are copied from the first range rather than the second, and that the relative order of elements in the output range is the same as in the first input range.

The two versions of set_intersection differ in how they define whether one element is less than another. The first version compares objects using operator<, and the second compares objects using a function object comp.

Definition

Defined in the standard header algorithm, and in the nonstandard backward-compatibility header algo.h.

Requirements on types

For the first version:

InputIterator1 is a model of Input Iterator.
InputIterator2 is a model of Input Iterator.
OutputIterator is a model of Output Iterator.
InputIterator1 and InputIterator2 have the same value type.
InputIterator's value type is a model of LessThan Comparable.
The ordering on objects of InputIterator1's value type is a strict weak ordering, as defined in the LessThan Comparable requirements.
InputIterator's value type is convertible to a type in OutputIterator's set of value types.

For the second version:

InputIterator1 is a model of Input Iterator.
InputIterator2 is a model of Input Iterator.
OutputIterator is a model of Output Iterator.
StrictWeakOrdering is a model of Strict Weak Ordering.
InputIterator1 and InputIterator2 have the same value type.
InputIterator1's value type is convertible to StrictWeakOrdering's argument type.
InputIterator's value type is convertible to a type in OutputIterator's set of value types.

Preconditions

For the first version:

[first1, last1) is a valid range.
[first2, last2) is a valid range.
[first1, last1) is ordered in ascending order according to operator<. That is, for every pair of iterators i and j in [first1, last1) such that i precedes j, *j < *i is false.
[first2, last2) is ordered in ascending order according to operator<. That is, for every pair of iterators i and j in [first2, last2) such that i precedes j, *j < *i is false.
There is enough space to hold all of the elements being copied. More formally, the requirement is that [result, result + n) is a valid range, where n is the number of elements in the intersection of the two input ranges.
[first1, last1) and [result, result + n) do not overlap.
[first2, last2) and [result, result + n) do not overlap.

For the second version:

[first1, last1) is a valid range.
[first2, last2) is a valid range.
[first1, last1) is ordered in ascending order according to comp. That is, for every pair of iterators i and j in [first1, last1) such that i precedes j, comp(*j, *i) is false.
[first2, last2) is ordered in ascending order according to comp. That is, for every pair of iterators i and j in [first2, last2) such that i precedes j, comp(*j, *i) is false.
There is enough space to hold all of the elements being copied. More formally, the requirement is that [result, result + n) is a valid range, where n is the number of elements in the intersection of the two input ranges.
[first1, last1) and [result, result + n) do not overlap.
[first2, last2) and [result, result + n) do not overlap.

Complexity

Linear. Zero comparisons if either [first1, last1) or [first2, last2) is empty, otherwise at most 2 * ((last1 - first1) + (last2 - first2)) - 1 comparisons.

Example

inline bool lt_nocase(char c1, char c2) { return tolower(c1) < tolower(c2); }

int main()
{
  int A1[] = {1, 3, 5, 7, 9, 11};
  int A2[] = {1, 1, 2, 3, 5, 8, 13};  
  char A3[] = {'a', 'b', 'b', 'B', 'B', 'f', 'h', 'H'};
  char A4[] = {'A', 'B', 'B', 'C', 'D', 'F', 'F', 'H' };

  const int N1 = sizeof(A1) / sizeof(int);
  const int N2 = sizeof(A2) / sizeof(int); 
  const int N3 = sizeof(A3);
  const int N4 = sizeof(A4);

  cout << "Intersection of A1 and A2: ";
  set_intersection(A1, A1 + N1, A2, A2 + N2,
                   ostream_iterator<int>(cout, " "));
  cout << endl 
       << "Intersection of A3 and A4: ";
  set_intersection(A3, A3 + N3, A4, A4 + N4, 
                   ostream_iterator<char>(cout, " "),
                   lt_nocase);
  cout << endl;
}

The output is

Intersection of A1 and A2: 1 3 5 
Intersection of A3 and A4: a b b f h

Notes

[1] Even this is not a completely precise description, because the ordering by which the input ranges are sorted is permitted to be a strict weak ordering that is not a total ordering: there might be values x and y that are equivalent (that is, neither x < y nor y < x) but not equal. See the LessThan Comparable requirements for a fuller discussion. The output range consists of those elements from [first1, last1) for which equivalent elements exist in [first2, last2). Specifically, if the range [first1, last1) contains n elements that are equivalent to each other and the range [first1, last1) contains m elements from that equivalence class (where either m or n may be zero), then the output range contains the first min(m, n) of these elements from [first1, last1). Note that this precision is only important if elements can be equivalent but not equal. If you're using a total ordering (if you're using strcmp, for example, or if you're using ordinary arithmetic comparison on integers), then you can ignore this technical distinction: for a total ordering, equality and equivalence are the same.