# [Dynamic Prog] Number Of Bus-Stations

At a bus-station, you have time-table for buses arrival and departure. You need to find the minimum number of platforms so that all the buses can be accommodated as per their schedule.

**Example:** Time table is like below:

BusArrivalDepartureBusA 0900 hrs 0930 hrs BusB 0915 hrs 1300 hrs BusC 1030 hrs 1100 hrs BusD 1045 hrs 1145 hrs

Then the answer must be 3. Otherwise the bus-station will not be able to accommodate all the buses.

**Answer:**

Let’s take the same example as described above. Now if we apply dynamic programming and calculate the number of buses at station at any time (when a bus comes or leaves). Maximum number in that pool will be nothing but the maximum number of buses at the bus-station at any time, which is same as max number of platforms required.

So first sort all the arrival(A) and departure(D) time in an int array. Please save the corresponding arrival or departure in the array also. Either you can use a particular bit for this purpose or make a structure. After sorting our array will look like this:

0900 0915 1930 1030 1045 1100 1145 1300 A A D A A D D D

Now modify the array as put 1 where you see A and -1 where you see D. So new array will be like this:

1 1 -1 1 1 -1 -1 -1

And finally make a cumulative array out of this:

1 2 1 2 3 2 1 0

Your solution will be the maximum value in this array. Here it is 3.

I think that code for this will not be complex so I am skipping that part. The complexity of this solution depends on the complexity of sorting.

**PS:** If you have a arriving and another departing at same time then put departure time first in sorted array.

**Further Thoughts:**You don not need to create a cumulative array or an array with 1 and -1; you just need a counter (cnt) initialized at '0'. Whenever, you find an 'A' in arrival-departure array, increment cnt by 1. Compare it with maximum value (max); if it is greater than max, make max equal to cnt. If you get a 'D' in arrival-departure array, decrement cnt by 1. At the end, return 'max'.