yeah interval tree and binary indexed tree is a one solution which will
give you answer in log(n) time for each query ,but if i got all the
interval at the beigning of time i can get solution in O(1) time by O(n
) preprocessing take array f initialize with zero,now for each
interval(l,r) do f[l]++ and f[r+1]-- once you are done wi all queries take
prefix sum value of each index will tell you in how many interval it was
On Sunday, June 9, 2013 12:50:46 PM UTC+5:30, Monish Gupta wrote:
>
> There are n Intervals. Given a set of m numbers, tell in how many
> intervals does each number exists.
> Example: 4 Intervals: [2,3] [3,4] [1,4] [3,10]. Numbers {2,4,11}
> For 2 -> 1 as 2 lies only in 1 interval [2,3]
> For 4 -> 3 as 4 lies in 3 intervals
> For 11 -> 0 as 11 lies in none of the given 4 intervals.
>
> It can be easily done in O(m*n) by traversing n intervals for each number
> in the given set of m numbers. How would improve this?
>
> Note: I could not fully recall, but I have seen very similar question in
> codechef but could not find the same.
>
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