Finding number of overlapping ranges for a given input range

Asked Mar 27 '17 at 16:15

Active Mar 28 '17 at 05:32

Viewed 96 times

What I have as input is a Range[] arr of size n where Range is

class Range{
    int upper, lower;
    Range(int upper,int lower){
        this.upper=upper;
        this.lower=lower;
    }

}

Now, for q queries, I get a Range r as input and I have to report the number of values in arr from index L to R(which is also given as input for each query),where a Range in arr is considered valid only if it's upper value is greater than r.upper and lower is less than r.lower.

I was thinking of using a merge sort tree, but I cannot figure out how to integrate both the conditions. Please suggest some data structure for which construction complexity as well as query complexity is O(n.log n) and O(q.log n) respectively.

Also, it should be noted that only the ranges within a particular segment is considered, which is also given as input.

PS: Some modification on merge sort tree or segment tree seems probable, so any answer with merge sort tree or segment tree is welcome.

EDIT : R[] = { {3,7} , {5,9} , {2,10} , {4,10} , {3,5} }

query #1 : {4,6}, l = 2 , r = 5

here, the ranges considered are {5,9} , {2,10} , {4,10} ,{3,5}

Out of these, only {5,9} , {2,10} , {4,10} are valid. So answer for this query is 3.

edited Mar 28 '17 at 05:32

asked Mar 27 '17 at 16:15

Rajarshi basu

1

can you give an example testcase? – Kaidul Mar 27 '17 at 16:31
Is it complete overlapping or partial overlapping also counts, for eg. for R[] = [{1,2}, {1,10}, {5, 9}] for query Q {6,10} , what is the expected answer ? – Amit Kumar Mar 27 '17 at 18:49
It may be possible to solve the task in O(sqrt(n) * n) if you can process queries offline with Mo's algorithm. – Ivan Gritsenko Mar 27 '17 at 19:07
@amit kumar No, partial overlapping does not count – Rajarshi basu Mar 28 '17 at 05:24

Finding number of overlapping ranges for a given input range

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