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  1. PO
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    <p>You could use <code>set_intersection</code> and test if the resulting set is empty, but I don't know if that's much faster. </p> <p>The optimal implementation would stop the testing and <code>return false</code> as soon as the first equal element is found. I don't know of any ready-made solution for that, though</p> <pre><code>template&lt;class Set1, class Set2&gt; bool is_disjoint(const Set1 &amp;set1, const Set2 &amp;set2) { Set1::const_iterator it, itEnd = set1.end(); for (it = set1.begin(); it != itEnd; ++it) if (set2.count(*it)) return false; return true; } </code></pre> <p>isn't too complex and should do the job nicely.</p> <p><strong>EDIT:</strong> If you want O(n) performance, use the <em>slightly</em> less compact</p> <pre><code>template&lt;class Set1, class Set2&gt; bool is_disjoint(const Set1 &amp;set1, const Set2 &amp;set2) { Set1::const_iterator it1 = set1.begin(), it1End = set1.end(); if (it1 == it1End) return true; // first set empty =&gt; sets are disjoint Set2::const_iterator it2 = set2.begin(), it2End = set2.end(); if (it2 == it2End) return true; // second set empty =&gt; sets are disjoint // first optimization: check if sets overlap (with O(1) complexity) Set1::const_iterator it1Last = it1End; if (*--it1Last &lt; *it2) return true; // all elements in set1 &lt; all elements in set2 Set2::const_iterator it2Last = it2End; if (*--it2Last &lt; *it1) return true; // all elements in set2 &lt; all elements in set1 // second optimization: begin scanning at the intersection point of the sets it1 = set1.lower_bound(*it2); if (it1 == it1End) return true; it2 = set2.lower_bound(*it1); if (it2 == it2End) return true; // scan the (remaining part of the) sets (with O(n) complexity) for(;;) { if (*it1 &lt; *it2) { if (++it1 == it1End) return true; } else if (*it2 &lt; *it1) { if (++it2 == it2End) return true; } else return false; } } </code></pre> <p>(modified Graphics Noob's modification further, using only operator &lt;)</p>
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    1. POc++ test if 2 sets are disjoint
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    1. PTAnswer
    1. UShjhill
    1. UShjhill
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    1. This table or related slice is empty.
    1. This table or related slice is empty.
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    1. This table or related slice is empty.
    1. VO
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      1. This table or related slice is empty.
      1. VTUpMod
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      1. VTUpMod
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      1. VTDownMod
    1. COActually, your implementation of is_disjoint is too complex (in the O(n) sense) - each iteration over set1 does an O(log(n)) lookup for the element in set2, giving a total runtime of O(nlog(n)), or O(nlog(n))/2 on average. This is possible in O(n) time.
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      1. USTerry Mahaffey
    2. COHow understandable something is is subjective. How it performs is not.
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      1. USTerry Mahaffey
    3. CO@Terry Mahaffey - yes, provided the iteration of both sets is possible in O(1) per step (I have to think about that, sets are usually implemented as some kind of tree structure). The code would be less easy to understand though...
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      1. UShjhill
 

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