Monish, please take a look at a similar problem about poor elephants in
the neighboring topic started today. I believe the problem has a similar
solution. Each start point increases the no. of active intervals by one;
each end point decreases it. So, we do the following:
1. Convert the input array of pairs (BTW, the points don't have to be of an
integral type!) into the following vector:
{ {2,+1}, {3,-1}, {3, +1}, {4, -1}...} This takes O(N) time and O(N)
additional memory
2. Sort the vector by: a) ascending .first; b) descending .second -- this
makes sure we count the intervals properly in case multiple intervals start
and end at the same point. Complexity is O(N*logN).
3. Convert "relative" counters to "absolute"; complexity is O(N).
int count = 0;
for (int i = 0; i < v.size(); ++i) {
count += v.second;
assert (count >=0);
v.second = count;
}
assert (count == 0);
4. After this preparation work ( O(N*logN) time, O(N) memory), answering
the question about the number of intervals for any numbers takes O(logN) --
use lower_bound() to binary search in the sorted vector.
On Sunday, June 9, 2013 3:20:46 AM UTC-4, 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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