There are n spectators in the stadium, labeled from 1 to n.
At time 1, the first spectator stands.
At time 2, the second spectator stands.
At time k, the k-th spectator stands.
At time k + 1, the (k + 1)-th spectator stands and the first spectator sits.
At time k + 2, the (k + 2)-th spectator stands and the second spectator sits.
At time n, the n-th spectator stands and the (n – k)-th spectator sits.
At time n + 1, the (n + 1 – k)-th spectator sits.
At time n + k, the n-th spectator sits.
Find the number of spectators standing in the stadium at a time t.
Input : 10 5 3 Output : 3 Explanation : n = 10, k = 5, t = 3 At time 1, 1st spectator stands. At time 2, 2nd spectator stands. At time 3, 3rd spectator stands. Thus, the result is 3 as there are total of 3 spectators standing. Input :10 5 7 Output : 5 Explanation : n = 10, k = 5, t = 7 At time 1, 1st spectator stands. At time 2, 2nd spectator stands. At time 3, 3rd spectator stands. At time 4, 4th spectator stands. At time 5, 5th spectator stands. At time 6, 6th spectator stands and 1st spectator sits [(n-k) = 6 - 5 = 1 as mentioned above]. At time 7, 7th spectator stands and 2nd spectator sits. So, now there are total of 5 spectators standing. Thus, result is 5.
Now we can observe certain conditions in this problem :
NOTE : There can only be maximum of k spectators standing in the stadium.
1) If the time is less than k value then that means spectators are still standing. So the result would be t.
2) In time between n and k(inclusive), the spectators standing are k. [Observation]
3) After n time, the spectators starts to sit one by one. So we calculate the spectators sitting by ‘t – n’. Then subtract k by this value. This produces the result.
Below is the implementation of the above approach :
Time Complexity : O(1)
Space Complexity : O(1)
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