Comments by nirvikprantor (Top 7 by date)

In this description the main point is...If we consider the planes delay in the run way above the example,where 2 planes are in each run way then: 1 2 2 3 3
1 1 1
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At 1 sec-1st,plane of the 1st,run way which delay time is 1sec,can flay from the run way.so now 1st plane is gone,so on this run way the second plane came in to the 1st position.
same as,in the second run way at 1 sec 1st plane can't take out from the run way...cause its delay time is 2.so at 1sec the plane can go to the middle of the run way..it need another 1 sec to take off from the run way.
That's how we need to find out the total delay time,what total time need to take off all the planes from the run way..an it will be calculate at,(1sec+1sec+1sec+....)=3

1st plane- 1 2 2 3 3
2nd plane- 1 1 1
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1sec- 0 1 1 2 2
1sec- 0 0 0 1 1
1sec- 0 0 0 0 0
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Total delay=(1+1+1)sec=3 sec