algorithm - Check if a number exists in a given set of ranges -

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Assume that a set of our N Series, (A1, B1), (A2, B2) (A3, B3) .. (A, BN), where A indicates the starting point and BE represents the end point of a boundary. (AI, BI positive Intijrs are)

How do we use binary search, if a given integer, X say, N present in at least one N range of?

My approach:

  1. Sort the ranges by the x coordinate and then y-coordinate.

  2. The smallest X-coordinate is bigger or equal to X.

  3. Check if the range is satisfied or not.

  4. If yes, we have a solution.

    Now, if there is no X in that category, what should my next step be?

    Or, should the solution be completely different?

    What is the description of your problem is that your pairs as (a1, b1) a), (A2, B-2) where is AX at the beginning of class bx finished now you are given a number Is n and you want to find that the number is in any category.
    sort before then implement the merger in overlapping categories and binary search: 

      #include & lt; Iostream & gt; # Include & lt; Algorithm & gt; # Include & lt; Vector & gt; using namespace std; Int main () {Vector & lt; Pair & lt; Int, int & gt; & Gt; B; B.push_back (make_pair (5,10)); B.push_back (make_pair (75,100)); B.push_back (make_pair (33,67)); B.push_back (make_pair (9, 21)); B.push_back (make_pair (28.667)); Int m = b.size (); Sort (B.Bigin (), B.End ()); Int j = 0; For (Int i = 1; i  i ++) {if (b [i]. First & lt; = b [ja] .second) {if (b [i] .second & gt; B [j] .second) {b [ja] .second = b [i] .second; }} And {j ++; B [j] = b [ii]; }} I = J + 1; {Cout & lt; for int i = 0; i & lt; m; i ++) & Lt; B [i]. First & lt; & Lt; "" & Lt; & Lt; B [i] .second & lt; & Lt; Endl; } // Apply binary search now; }   
     I hope this will solve your problem. I left the binary search section as exercise for you.

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