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Home/ Questions/Q 473703
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T00:12:49+00:00 2026-05-13T00:12:49+00:00

The standard way of intersecting two sets in C++ is to do the following:

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The standard way of intersecting two sets in C++ is to do the following:

std::set<int> set_1;  // With some elements
std::set<int> set_2;  // With some other elements
std::set<int> the_intersection;  // Destination of intersect
std::set_intersection(set_1.begin(), set_1.end(), set_2.begin(), set_2.end(), std::inserter(the_intersection, the_intersection.end()));

How would I go about doing an in-place set intersection? That is, I want set_1 to have the results of the call to set_intersection. Obviously, I can just do a set_1.swap(the_intersection), but this is a lot less efficient than intersecting in-place.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-13T00:12:49+00:00Added an answer on May 13, 2026 at 12:12 am

    I think I’ve got it:

    std::set<int>::iterator it1 = set_1.begin();
    std::set<int>::iterator it2 = set_2.begin();
    while ( (it1 != set_1.end()) && (it2 != set_2.end()) ) {
        if (*it1 < *it2) {
            set_1.erase(it1++);
        } else if (*it2 < *it1) {
            ++it2;
        } else { // *it1 == *it2
                ++it1;
                ++it2;
        }
    }
    // Anything left in set_1 from here on did not appear in set_2,
    // so we remove it.
    set_1.erase(it1, set_1.end());
    

    Anyone see any problems? Seems to be O(n) on the size of the two sets. According to cplusplus.com, std::set erase(position) is amortized constant while erase(first,last) is O(log n).

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