Efficient data structure for finding non-overlapping range containing a number?

(I have changed this answer since the asker clarified that his ranges do not overlap. ).

(I have changed this answer since the asker clarified that his ranges do not overlap. ) If the set of ranges does not change, you can use a sorted vector and binary search, as suggested in ravenspoint's answer. If the set of ranges changes over time, you might still use a sorted vector, or you might want to use a std::map.

You need to try both and see which one is faster in that case.

I still find problems with actually finding whether an element is present in a range. The problem is it should be found either its present in anyone of the range strictly or not?! No mix and matching the ranges.

– Dumb yesterday.

Assuming the ranges do not overlap: Store each range in a simple structure range { int low; int high; string name; } Store the ranges in a sorted vector, by low. Find required range using binary search for largest low less than target.

S range1 and range3 examples overlap. – rob mayoff Dec 14 at 21:35 Apparently they do not overlap. The question is confusing.

I have edited question so example does not overlap – ravenspoint Dec 15 at 13:59 Sorry for the mistake in the question. You have edited before I could. I dint notice the overlap.

– Dumb Dec 15 at 21:51.

Just dump all the values , starting and ending into a vector or array, and then sort it. Since the ranges don't overlap, once the array is sorted, you will have start, stop,start,stop,etc.. then, you could use a binary search to find the index of the vector. Then its just a question of whether its odd or even assuming, you are getting the ranges from the stream vector ranges; int n; while(in >> n){ ranges.

Push_back(n); } sort(ranges.begin(),ranges.end()) int x; cout > x; int index = binary_search(x,ranges); if(index % 2){ cout = f) return s; int n = (f-s)/2 + s; if(vecn == x)return n; if(vecn.

I need to find whether the element is present in a range and it should not mix and matched and it should be strictly either present or not. – Dumb yesterday.

With B+ tree, the fan-out would be small and search would be fast too.

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