how to find the empty spot in a row? - combinatorics

Question

I am suppose to put a group of people (given at run-time) in to a 2 dimensional array of spot, group together in a row randomly(find out all possible positions and randomly pick one)
To start of, I wanna try in an array first
if I have an array of spots of size 10 like below
spots[occupied,open,open,occupied,occupied,occupied,open,open,open,open]
to put 2 people
Is there a particular algorithm to do this kind of problem?
Thanks for any help!

Answer 1

in python:
seats =["occupied","open","open","occupied","occupied","occupied","open","open","open","open"]
def empty(seats,index,count):
if count == 0:
return True
else:
return (seats[index] == "open") & empty(seats,index+1,count-1)
def findEmpty(seats,count):
result = []
for (i=0;i<seats.size-count+1,i++)
if empty(seats,i,count):
result.append(<list of consecutive numbers from i to i+count>)
return result
print findEmpty(seats,2)
>>>[[1, 2], [6, 7], [7, 8], [8, 9]]
here's another approach, its a little more efficient:
seats = ["occupied","open","open","occupied","occupied","occupied","open","open","open","open"]
//counts the number of consecutive free seats starting at position index
def countEmpty(seats,index):
if index >= len(seats) or seats[index] == "occupied":
return 0
return 1 + countEmpty(seats,index+1)
def findEmpty(seats,count):
result = []
i = 0
while i < len(seats)-count+1:
c = countEmpty(seats,i)
if c>=count:
for (j=i;j<i+c-count+1;j++):
result.append(<list of consecutive numbers from j to j+count>)
i += 1 + c
return result
print findEmpty(seats,2)
>>>[[1, 2], [6, 7], [7, 8], [8, 9]]
and finally, if you did choose to use python you could do it in one line:
seats =["occupied","open","open","occupied","occupied","occupied","open","open","open","open"]
count = 2
print [range(i,i+count) for i in range(len(seats)-count+1) if all([seats[j]=="open" for j in range(i,i+count)]) ]
>>> [[1, 2], [6, 7], [7, 8], [8, 9]]

Answer 2

Pseudo code:
//Create 2 lists of blocks, both empty: tmp and final
List tmp=new List;
List final=new List;
//Cycle the seats
for (int i=0;i<length(seats);i++) {
//If the seat is occupied, start over
if (seats[i]==occupied) tmp.empty();
else {
//Cycle existing block candidates, add free seat
foreach (ref block in tmp) {
block.add(seats[i])
if (length(block)>=people_count) {
//HEUREKA, got a fitting block: Move it to the final list
tmp.remove(block)
final.add(block)
}
}
//Start a new block with this seat
tmp.add(new block(seats[i]));
//Read below for this edge case
}
}
final now has the blocks.
If you allow the edge case of people_num being 1, you have to check for a complete block at the position indicated in the pseudocode

Answer 3

I shall use Mathematica code but I believe you can follow the logic.
Starting with:
dat = spots[occupied, open, open, occupied, occupied, occupied, open, open, open, open];
fill = {"Alpha", "Bravo", "Charlie", "Delta", "Echo", "Foxtrot"};
First find a random permutation of the list fill (it is easy to find an algorithm on StackOverflow):
randomfill = RandomSample # fill
{"Delta", "Echo", "Alpha", "Bravo", "Charlie", "Foxtrot"}
Then "map" a function onto each element of the spots list, and if the element is open return the next value from the randomfill list, else return the element unchanged:
i = 1;
If[# === open, randomfill[[i++]], #] & /# dat
spots[occupied, "Delta", "Echo", occupied, occupied, occupied, "Alpha", "Bravo", "Charlie", "Foxtrot"]