Theatre Seating

A theatre has 10 rows, labelled A to J from front to back, and 15 seats in each row, numbered 1 to 15 from left to right.

The program uses the following rules to choose the best seats.

• Rule 1: All seats in one booking must be in the same row, next to each other.
• Rule 2: The seats must be as close to the front as possible, then as close to the left as possible (lowest letter, then lowest number)

Write a function which takes the number of tickets wanted as an integer input (n), and outputs the best seats available in a list of length n.

• Output -1 if 1 > Input or Input > 15*
• Output -1 if the seats aren't available*
• Have a function B(n) that the user can use to input the desired number of seats.

*You can output the -1 in a list if it makes it easier

Examples

I/O

Calling B(5) on a new array should return [A1, A2, A3, A4, A5]
Calling B(2) after that should then return [A6, A7]
Calling B(10) after that should then return [B1, B2, ... B9, B10]
Calling B(-1) should always return -1

Un-golfed Solution Python

Theatre = [ [False] * 16 ] * 11

def B(n):
if 0 <= n <= 15:
for i in range(10):
for j in range(15-n+1):
try:
if not Theatre[i][j]:
if not Theatre[i][j + n]:
row = i
start = j
List = []
for q in range(n):
List.append(chr(row + 65) + str(start + q + 1))
Theatre[row][start + q] = True
return List
except:
break
return -1

• Is "Have hardcoded a list of seats in a two dimensional array" necessary? There are numerous ways to do this without that; the requirement really restricts solutions. Commented May 23, 2014 at 1:42
• You say the 2-D array must be hard-coded, but your Python example doesn't even hard-code it, it uses a comprehension to create a new list at runtime. Commented May 23, 2014 at 1:53
• Why even mention "a list of seats in a two dimensional array"? That sounds like an implementation detail and if somebody creates a program that satisfies the required output without using an array, there should be no problem with that. Commented May 23, 2014 at 2:19
• what if input is 0? Commented May 23, 2014 at 15:17
• @edc65 I always make my nonexistent movie theater patrons sit in the best spot of the theater, on another patron's lap if need be. They never notice. Commented May 23, 2014 at 15:56

JavaScript - 172

Function itself is 172:

//build persistent seats
m=[];
for(i=10;i--;){m[i]={r:String.fromCharCode(i+65),s:[]};for(j=0;j<15;j++)m[i].s.push(j+1);}

function b(z){for(i=0;i<m.length;i++)for(j=0,u=m[i].s.length;o=[],j<u;j++)if(u>=z&z>0){for(m[i].s=m[i].s.slice(z),p=m[i].s[0]||16;o[--z]=m[i].r+--p,z;);return o;}return-1;}


Input:

console.log(b(-1));
console.log(b(0));
console.log(b(4));
console.log(b(15));
console.log(b(1));
console.log(b(20));


Output:

-1
-1
[ 'A1', 'A2', 'A3', 'A4' ]
[ 'B1', 'B2', 'B3', 'B4', 'B5', 'B6', 'B7', 'B8', 'B9', 'B10', 'B11', 'B12', 'B13', 'B14', 'B15' ]
[ 'A5' ]
-1


Javascript (ES6) - 130127107101 98

B=n=>(a=>{for(;n>0&a<9;)if((b=~~B[++a]+n)<16)for(B[a]=b;n--;)c[n]='ABCDEFGHIJ'[a]+b--})(c=[-1])||c


Demo here: http://jsfiddle.net/tBu5G/

Some ideas taken from @edc65

• c[B[a]=b] instead of c[],B[a]=b is clever, but fails for n=0 Commented May 24, 2014 at 7:25
• @edc65 nice catch. I've now adjusted it to handle the case n=0 Commented May 24, 2014 at 15:16
• Awesome. That's something to remember to avoid 'return' - thanks for sharing (+1) Commented May 25, 2014 at 21:41
• @edc65 thanks! I thought it was interesting. MT0's got us both beat though! :P Commented May 25, 2014 at 21:45

t=[[a:show s|s<-[1..15]]|a<-['A'..'J']]
b n=(n%).span((<n).length)
_%(h,[])=([],h)
n%(j,(r:s))=let(t,u)=splitAt n r in(t,j++u:s)


Some adjustments had to be made to make this a function in Haskell: b returns a pair: the tickets (if possible), and the new state of the theater. t is the initial theater state, with all tickets unsold. Also, returning -1 was unnatural for Haskell, so if no tickets can be issued for a request, the empty list is returned for the tickets.

λ: let (k1,t1) = b 5 t
λ: k1
["A1","A2","A3","A4","A5"]

λ: let (k2,t2) = b 2 t1
λ: k2
["A6","A7"]

λ: let (k3,t3) = b 10 t2
λ: k3
["B1","B2","B3","B4","B5","B6","B7","B8","B9","B10"]

λ: let (k4,t4) = b (-1) t3
λ: k4
[]

λ: let (k5,t5) = b 2 t4
λ: k5
["A8","A9"]


APL (75)

T←10 15⍴0⋄B←{(⍵∊⍳15)∧∨/Z←,T⍷⍨⍵/0:+T[P]←{⎕A[⍺],⍕⍵}/¨P←(⊃Z/,⍳⍴T)∘+¨1-⍨⍳1⍵⋄¯1}


Test:

      B 5
A1    A2    A3    A4    A5
B 2
A6    A7
B 10
B1    B2    B3    B4    B5    B6    B7    B8    B9    B10
B ¯1
¯1
B 3
A8    A9    A10


Explanation:

• T←10 15⍴0: T is a 15-by-10 matrix that holds the theater state (0 = free)
• B←{...}: the function
• (⍵∊⍳15): if ⍵ is a member of the set of integers from 1 to 15,
• ∨/Z←,T⍷⍨⍵/0: and T contains ⍵ zeroes in a row (storing possible start points in Z),
• :: then:
• (⊃Z/,⍳⍴T): select possible start coordinates, and take the first one,
• ∘+¨1-⍨⍳1⍵: add ⍵-1 more positions to the right of the start coordinate
• P←: store the coordinates in P
• {⎕A[⍺],⍕⍵}/¨: format the coordinates
• T[P]←: store the formatted coordinates at their places in T. (any nonzero values in T will do)
• +: return the result, which is the formatted coordinates (the result of an assignment is tacit by default)
• ⋄¯1: otherwise, return ¯1.

Javascript (E6) 99 103 113 121

Really you just need to store a number for each row

B=n=>{for(r=i=[-1];n>0&i++<9;)if((a=~~B[i]+n)<16)for(B[i]=a;n--;)r[n]='ABCDEFGHIJ'[i]+a--;return r}


Test

'5:'+B(5)+'\n2:'+B(2)+'\n10:'+B(10)+'\n0:'+B(0)+'\n1:'+B(-1))+'\n3:'+B(3)


Ungolfed

B = n => {
for (r = i = [-1]; n > 0 & i++ < 9;)
if ((a = ~~B[i] + n) < 16)
for (B[i] = a; n--; ) r[n] = 'ABCDEFGHIJ'[i] + a--;
return r;
}


JavaScript (ECMAScript 6 Draft) - 9695 91 Characters

A recursive solution:

Version 1

B=(n,r=0)=>n>0&&(k=~~B[r])+n<16?[...Array(n)].map(_=>'ABCDEFGHIJ'[r]+(B[r]=++k)):r<9?B(n,r+1):-1


Version 2:

B=(n,r=0)=>n<1|r>9?-1:(k=B[r]|0)+n<16?[...Array(n)].map(_=>'ABCDEFGHIJ'[r]+(B[r]=++k)):B(n,r+1)


(Thanks to nderscore for the inspiration for the 1 character saving)

Version 3:

B=(n,r=0)=>n<1|r>9?-1:(B[r]^=0)+n<16?[...Array(n)].map(_=>'ABCDEFGHIJ'[r]+ ++B[r]):B(n,r+1)


(Thanks to nderscore)

Explanation:

B = function(n,r=0)          // Create a function B with arguments:
// - n is the number of seats to book
// - r is the row number (defaults to 0)
{
var k = ~~B[r];            // get the number of seats already booked in row r
if (  n > 0                // ensure that n is a valid booking
&& k+n<16 )             // check that there are enough seats remaining in row r
{
var P = new Array(n);    // Create an array with length n with no elements initialised
var Q = [...P];          // Use P to create an array with every element
// initialised to undefined
var R = 'ABCDEFGHIJ'[r]; // get the row ID.
B[r] = k + n;            // Increment the number of seats booked in row r by n.
var S = Q.map(
function(){
return R + (++k);    // Map each value of Q to the row ID concatenated with
// the seat number.
}
);
return S;                // Return the array of seats.
}
else if ( r < 9 )          // If there are more rows to check
{
return B(n,r+1);         // Check the next row.
}
else                       // Else (if n is invalid or we've run out of rows)
{
return -1;               // Return -1.
}
}

• Nice solution. I was working on something similar. Here's -1 byte: B=(n,r=0)=>n>0&r<9?(k=B[r]|0)+n<16?[...Array(n)].map(_=>'ABCDEFGHIJ'[r]+(B[r]=++k)):B(n,r+1):-1 Commented May 25, 2014 at 21:42
• Thanks, unfortunately that one doesn't quite work as you can't book row J but negating the first check to give B=(n,r=0)=>n<1|r>9?-1:(k=B[r]|0)+n<16?[...Array(n)].map(_=>'ABCDEFGHIJ'[r]+(B[r]=++k)):B(n,r+1) should work.
– MT0
Commented May 25, 2014 at 22:16
• Ah, good catch. Commented May 25, 2014 at 22:24
• And it keeps going lower... (91) B=(n,r=0)=>n<1|r>9?-1:(B[r]^=0)+n<16?[...Array(n)].map(_=>'ABCDEFGHIJ'[r]+ ++B[r]):B(n,r+1) Commented May 26, 2014 at 6:59

GolfScript, 103 82 bytes

226,1>15/[0]*:T{:&0>{T[{),&~)>:|T\/,2=}?]{T|-:T;|{(.[15/65+]\15%)+}%}-1if}-1if}:B


Examples

$cat theatre.gs 226,1>15/[0]*:T {:&0>{T[{),&~)>:|T\/,2=}?]{T|-:T;|{(.[15/65+]\15%)+}%}-1if}-1if}:B 5 B p # Execute B(5), stringify and print. 2 B p 15 B p 17 B p 0 B p {}:puts # Disable automatic output.$
$golfscript theatre.gs ["A1" "A2" "A3" "A4" "A5"] ["A6" "A7"] ["B1" "B2" "B3" "B4" "B5" "B6" "B7" "B8" "B9" "B10" "B11" "B12" "B13" "B14" "B15"] -1 -1  How it works 226,1> # Push the array [ 1 … 225 ]. 15/[0]* # Split in chunks of 15 elements and join separating by zeros. :T # Save result in T. { # :&0> # Save the function's argument in & and check if it's positive. { # If it is: T[{ # For each seat S in T: ), # Push [ 0 … S ]. &~)> # Reduce two [ S-(&-1) … S ]. :| # Save the result in |. T\/ # Split T around |. ,2= # If there are two chunks, the seats are available. }?] # Find the first S that satisfies the above condition. { # If there was a match: T|-:T; # Remove the seats in | from T. |{ # For each seat S in |: (. # Push S+1 S+1. [15/65+] # Compute (S+1)/15+65; the ASCII character corresponding to the row. \15%)+ # Compute (S+1)%15+1, stringify and concatenate. }% # } # -1if # If there was no match, push -1 instead. } # -1if # If the argument was non-positive, push -1 instead. }  CoffeeScript - 171150 149 I suspect Ruby or Perl will beat this out before long. c=0;l=64;k=1 f=(n)-> if n<0 or n>15 or 150-c<n return-1 a=[] for i in[1..n] if c%15==0 ++l;k=1 ++c;a.push String.fromCharCode(l)+k;++k a  Equivalent JavaScript/Explanation: For those unfamiliar with CoffeeScript. var seats = 0; //Occupied seats. var letter = 64; //ASCII code for row letter. var index = 1; //Index of seat in row. function seats( count ) { if( count < 0 || count > 15 || ( 150 - seats ) < count ) return -1; var assignedSeats = []; for( var i = 1; i <= count; ++i ) { if( ( seats % 15 ) === 0 ) { ++letter; index = 1; } ++seats; //Occupy a seat. assignedSeats.push( String.fromCharCode( letter ) + index ); ++index; } return assignedSeats; }  • This solution doesn't satisfy the rule All seats in one booking must be in the same row, next to each other. Commented May 23, 2014 at 4:26 Cobra - 309 This should do it, but I can't actually get to a compiler for a few hours, so I'll update it later if needed. class P var s=List<of List<of String>>() def main for l in 'ABCDEFGHIJ' t=[] for n in 1:16,t.insert(0,l.toString+n.toString) .s.add(t) def b(n) as List<of String> t=[] for r in .s.count,if .s[r].count>=n for i in n,t.add(.s[r].pop) break return if(n>0 and t<>[],t,['-1'])  C# - 289 First attempt at code golfing. int[]s=new int[10];string[]B(int n){string[]x=new string[]{"-1"};if(n<1||n>15)return x;int m=(int)Math.Pow(2, n)-1;for(int i=0;i<10;++i){for(int j=0;j<15-n;++j){if((s[i] &m)==0){s[i]|=m;string[]r=new string[n];for(int k=0;k<n;++k)r[k]=(""+(char)(i+65)+(j+k+1));return r;}m<<=1;}}return x;}  Un-golfed int[] s = new int[10]; string[] B(int n) { string[] x = new string[] { "-1" }; if (n < 1 || n > 15) return x; int m = (int)Math.Pow(2, n) - 1; for (int i = 0; i < 10; ++i) { for (int j = 0; j < 15 - n; ++j) { if ((s[i] & m) == 0) { s[i] |= m; string[] r = new string[n]; for (int k = 0; k < n; ++k) r[k] = ("" + (char)(i + 65) + (j+k+1)); return r; } m <<= 1; } } return x; }  K, 140 d:10#,15#0b B:{if[(x<0)|x>15;:-1];$[^r:*&&/'~:^a:{(*&&/'{x(!1+(#x)-y)+\:!y}[d x;y])+!y}[;x]'!#d;-1;[.[d;(r;a r);~:];(10#.Q.A)[r],/:\$1+a r]]}


There are undoubtedly numerous improvements to be made here

C++ - 257

Also a first attempt at golfing.

static vector< int > t (10, 0);

vector<string> b(int n){
vector<string> o;
int i=0,j;
for(;i<10&&16>n&&n>0;i++){
if(15-t[i]<n) continue;
char l='A'+i;
for(j=t[i];j<n+t[i];j++){
o.push_back(l + toS(j + 1));
}
t[i]+=n;
n=0;
}
if(o.empty()) o.push_back("-1");
return o;
}


Because to_string wasn't working with my compiler, toS is defined as

string toS(int i){
return static_cast<ostringstream*>( &(ostringstream() << i) )->str();
}


And as a little interface

int main(){
int input = 0;
bool done = false;
while (!done){
cout << "how many seats would you like? (0 to exit)\n";
cin >> input;
vector<string> selection = b(input);
for (auto s : selection){
cout << s << ' ';
}
cout << endl;
if (input == 0) break;
}
return 0;
}

• Just removing unnecessary whitespace brings it down to 243 characters. Commented May 23, 2014 at 15:30
• More golfing to 236: vector<int> t(10,0);vector<string> b(int n){vector<string> o;for(int i=0,j;i<10&&16>n&&n>0;i++){if(15-t[i]<n)continue;char l='A'+i;for(j=0;j<n;j++)o.push_back(l+to_string(j+t[i]+1));t[i]+=n;n=0;}if(o.empty())o.push_back("-1");return o;} Commented May 23, 2014 at 15:38

C# - 268 Bytes

Golfed code:

int[]s=new int[10];string[]B(int n){string[]x={"-1"};if(n<1||n>15)return x;int m=(int)Math.Pow(2,n)-1;for(int i=0;++i<10;){for(int j=0;++j<15-n;){if((s[i]&m)==0){s[i]|=m;var r=new string[n];for(int k=0;++k<n;)r[k]=(""+(char)(i+65)+(j+k+1));return r;}m<<=1;}}return x;}


Ungolfed code:

    int[] s = new int[10];
string[] B(int n)
{
string[] x = { "-1" };
if (n < 1 || n > 15) return x;
int m = (int)Math.Pow(2, n) - 1;
for (int i = 0; ++i < 10;)
{
for (int j = 0; ++j < 15 - n;)
{
if ((s[i] & m) == 0)
{
s[i] |= m;
var r = new string[n];
for (int k = 0; ++k < n;)
r[k] = ("" + (char)(i + 65) + (j + k + 1));
return r;
}
m <<= 1;
}
}
return x;
}


I would have written some annotations into a comment on GoldenDragon's solution instead of making my own, but my reputation doesn't allow it.