Bus Station Puzzle -A DP Approach

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:


Bus


Arrival


Departure

BusA


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.

The complexity of this solution depends on the complexity of sorting.

Soln Provided By Aakash