Draw the Sawtooth Alphabet

Question

A simple one today. Write the shortest program that draws a "sawtooth alphabet" given a positive integer for the height. You must write the name of your programming language when you come to the letter it starts with.

For example, if your language is Python and the input is 1 the output should be:

ABCDEFGHIJKLMNOPythonQRSTUVWXYZ

If the input is 2 the output should be:

 B D F H J L N Python R T V X Z
A C E G I K M O      Q S U W Y

If the input is 4 the output should be:

   D     J     Python     V
  C E   I K   O      Q   U W
 B   F H   L N        R T   X Z
A     G     M          S     Y

Notes

• A always starts at the bottom left.
• Input can be via stdin or function call or similar. Output to stdout.
• Input above 26 doesn't need to work.
• No trailing spaces.

Answer 1

C, 134

n,m,c,p;
main(r){
for(scanf("%d",&n),m=--n?n*2:1;n+r--;puts(""))
for(c=-1,p=1;c++<25;)
p+=(c-n-r)%m*((c-n+r)%m)?1:1-printf("%*c",p,c+65);
}

Compiles on gcc with a few warnings. Newlines aren't included in the character count.

122 characters if the input is already stored in n.

Thanks to user2992539, tolos and edc65 for improvements.

Answer 2

if n holds the height:

C + escape codes: 81

x;main(y){for(y=n--;x<26;x++)printf("\033[%d;%dH%c",n?x/n&1?y++:y--:y,x+1,x+65);}

C: 110

x;char a[702]={[0 ...701]=32};main(y){for(y=--n;x<26;a[x*27-1]=10)a[27*(n?x/n&1?y++:y--:y)+x]=x+++65;puts(a);}

Answer 3

TI-BASIC - 148 bytes (raw text), 186 bytes (graphic)

In response to the OP, the wonderful TI-83 (and newer) comes with a size of 16 x 8 (just using the standard large text) or with a size of 94 x 62 pixels (which with small text is worth about 10 lines).

Now, this has a little problem (to which I would like clarification). The size impositions can't be "ignored" by the interpreter; in other words, if we were to try to set the sawtooth height at 20, it would give an error preventing the full execution of the code. I could write code that would, in an infinite environment, produce the correct output, except it wouldn't run on the machine.

With this being said, I present you the (running) versions of the programs. They all depend on the variable N being set to the desired height in lines before running:

• Raw text approach

  :ClrHome
  :"ABCDEFGHIJKLMNOPQRSTI-BASICUVWXYZ"→Str1
  :8→R
  :For(C,1,16
  :If N=1
  :Then
  :0→F
  :Else
  :If R<2 or 10-R>N
  :1→F
  :If R>7
  :-1→F
  :End
  :If C≠20
  :Then
  :Output(R,C,sub(Str1,C,1
  :Else
  :Output(R,C,sub(Str1,C,8
  :C+7→C
  :End
  :R+F→R
  :End
  

  In order to make this work regardless of the terminal, change For(C,1,16 to For(C,1,33 and remove the upper bound checking (R<2 or). Here is the output with 5→N:

  

• Graphic approach (this may also need AxisOff for clarity)

  :ClrDraw
  :N*6→N
  :"ABCDEFGHIJKLMNOPQRSTI-BASICUVWXYZ"→Str1
  :57→R
  :For(C,1,56
  :If N=6
  :Then
  :0→F
  :Else
  :If R<7 or 64-R>N
  :6→F
  :If R>51
  :-6→F
  :End
  :If C≠20
  :Then
  :If C>50
  :Then
  :Text(R,C,sub(Str1,C-23,1
  :Else
  :Text(R,C,sub(Str1,C,1
  :End
  :Else
  :Text(R,C,sub(Str1,C,8
  :C+30→C
  :End
  :R+F→R
  :End
  

  This one works OK, with two minor quirks. The height is still a problem, although the width isn't. However, I didn't space the letters, so in some cases (when the letter starts to rise or decline from the sawtooth), the letters may be chopped off by their successors. To make it work regardless of the terminal, remove the upper bound check (R<7 or). Then follows the graph:

  

Answer 4

Pure Bash (no coreutils), 181 bytes

m=$1
for l in A Bash {C..Z};{
((m))||s=++
((m>$1-2))&&s=--
for((m=$1==1?1:m,m$s,i=0;i<$1;i++));{
((i-m))&&a[i]+=${l//?/ }||a[i]+=$l
}
}
shopt -s extglob
printf %s\\n "${a[@]%%+( )}"

Output:

Piped to cat -E just to prove there are no trailing newlines.

$ ./sawtooth.sh 1 | cat -E
ABashCDEFGHIJKLMNOPQRSTUVWXYZ$
$ ./sawtooth.sh 5 | cat -E
       E       M       U$
      D F     L N     T V$
     C   G   K   O   S   W$
 Bash     H J     P R     X Z$
A          I       Q       Y$
$ 

Answer 5

JavaScript (ES6) 231 244

Edit Bug fix, some reordering and a different way of managing height==1
Moreover, changed to a function as this is allowed by OP, so no prompt() for input

No changes to the general algorithm that probably is NOT the best for this challenge

F=h=>{for(p=s=[i=z=b=t=''];++p<h;)i=1,z=b=' ',s[p]=t;for(p=--h,k=64;++k<91;)s[p]+=t+String.fromCharCode(k),k-74||(s=s.map((x,q)=>x+(p-q?'         ':'avascript'))),p-=i,p<0|p>h?(i=-i,p-=i+i,t=z,b=t+t):t+=b;console.log(s.join('\n'))}

Explained

F=h=>{
  // row in s for output must be initialized to ''. In the same step I make other initializations
  for(p=s=[i=z=b=t='']; ++p < h;) // initialize for height 1, all increment and spacing can be null
    i=1,z=b=' ',s[p]=t; // the for body is entered only if height > 1, initializing spacing and increment to the right values
  for(p=--h,k=64;++k<91;)
    s[p]+=t+String.fromCharCode(k),
    k-74||(s=s.map((x,q)=>x+(p-q?'         ':'avascript'))), // if 'J' modify each line of output adding the name or spacing
    p-=i,p<0|p>h?(i=-i,p-=i+i,t=z,b=t+t):t+=b; // index bouncing
  console.log(s.join('\n'))
}

Examples

1

ABCDEFGHIJavascriptKLMNOPQRSTUVWXYZ

3 going up

  C   G            K   O   S   W
 B D F H Javascript L N P R T V X Z
A   E   I            M   Q   U   Y

4 on top

   D     Javascript     P     V
  C E   I          K   O Q   U W
 B   F H            L N   R T   X Z
A     G              M     S     Y

7 going down

      G                    S
     F H                  R T
    E   I                Q   U
   D     Javascript     P     V
  C                K   O       W
 B                  L N         X Z
A                    M           Y

Answer 6

JAVA (393)

As always a great language for golfing:

public class P{public static void main(String[]a){int h=Integer.valueOf(a[0]);String x="ABCDEFGHIJKLMNOPQRSTUVWXYZ";String[]l=new String[h];Arrays.fill(l,"");int i=h-1;int d=-1;for(char c:x.toCharArray()){for(int n=0;n<l.length;n++){String k=(n==i)?(c=='J')?"Java":c+"":(c=='J')?"    ":" ";l[n]+=k;}if(i==0&&d==-1)d=1;if(i==h-1&&d==1)d=-1;if(h>1)i+=d;}for(String s:l){System.out.println(s);}}}

Answer 7

Ruby, 112 bytes

o="#{' '*29}
"*h
26.times{|i|o[(h-i%(h<2?1:2*h-2)-1).abs*30+i+(i>17?3:0)]=(i+65).chr}
puts o.gsub('R   ','Ruby')

Expects the input to be stored in h.

Let me know if this needs clarification.

Answer 8

J : 75 bytes

NB. without IO |:26([:u:65+i.@[)`(($(,|.@}.@}:)@i.)<"1@,.i.@[)`(' '$~,~)}5
2(1!:2)~|.26([:u:65+i.@[)`(($(,|.@}.@}:)@i.)<"1@,.i.@[)`(' '$~,~)}".1!:1[1

Using the wonderful Amend conjunction. As usual the IO is ugly and clunky, not going into details there. The core solution takes 3 gerunds (a sort of noun-ified verb (aka. function)):

• a) generating the alphabet
• b) generating the indices

• c) generating the matrix to make amends to

  x (abc}) y

a) is a pretty trivial lookup in the ascii table

c) is even more trivial

b) is the interesting one. The morale is that the horizontal indices are supposed to be starting from 0 going to y-1 and back down, repeating this 26 times. Eg. for y == 4:

0 1 2 3 2 1 0 1 2 3 2 1 ...

Implementing this gives for b):

($(,|.@}.@}:)@i.) <"1@,. i.@[)       NB. is equivalent too
(tine0 tine1 tine2)                  NB. a fork with tines defined as
tine0 =: 
  hook0 =: hook0_0 hook0_1           NB. a dyadic hook: x (g h) y -: x g h y
     hook0_0 =: $                    NB. reshape
     hook0_1 =: (hook1_0 hook1_1)@i. NB. do hook1 after making 0-y
       hook1_0=: ,                   NB. append to self
       hook1_1=: |.@}.@}:            NB. rotated version of y after beheading and curtailing
tine2 =: i.@[                  NB. generate 0-x
tine1 =: <"1@,.                NB. glue together coordinates.

And oh yea, handy fact: J's name is ... "J".

Answer 9

R (204)

f=function(x) {
  m=matrix(" ",x,26)
  i=x
  d=ifelse(x==1,0,-1)
  for (j in 1:26) {
    m[i,j]=LETTERS[j]
    if (!((i+d) %in% 1:x)) d=-d
    i=i+d
  }
  for (i in 1:x) cat(m[i,],"\n",sep="")
}

Result

> f(1)
ABCDEFGHIJKLMNOPQRSTUVWXYZ
> f(2)
 B D F H J L N P R T V X Z
A C E G I K M O Q S U W Y 
> f(3)
  C   G   K   O   S   W   
 B D F H J L N P R T V X Z
A   E   I   M   Q   U   Y 
> f(7)
      G           S       
     F H         R T      
    E   I       Q   U     
   D     J     P     V    
  C       K   O       W   
 B         L N         X Z
A           M           Y 

Answer 10

J, 67 57 61 characters

echo@dtb"1@|.@|:@(26&,$(u:65+i.26)#~1 j.26$(0#~1=]),<:#-&0 2)

Use as a function:

   echo@dtb"1@|.@|:@(26&,$(u:65+i.26)#~1 j.26$(0#~1=]),<:#-&0 2)
   D     J     P     V
  C E   I K   O Q   U W
 B   F H   L N   R T   X Z
A     G     M     S     Y

---

Explanation: this solution uses a different approach than the other J solution. Instead of generating a sawtooth wave 0 1 2 3 2 1 0 1 ..., I looked at the spacing between consecutive letters. For instance, for n=4 if you go from the A upwards, wrap over to the second column and reach B, then you find four spaces between A and B. This pattern of spacing between letters is very regular: for n=4 the pattern is 4 4 4 2 2 2 4 4 4 ....

So the idea is to first build the flattened (and transposed) array, and then re-shape it and flip it so it look right. The output routine is straightforward (for being J, at least): dtb is "delete trailing blanks" and "1 says "operate on each line". dtb and echo are both provided by the standard library.

Thanks to Zsbán Ambrus for some golfing help.

Answer 11

Javascript - 248 224 235

z=~~prompt(),y="ABCDEFGHIJKLMNOPQRSTUVWXYZ".split("");y[9]="Javascript";for(w=[],v=0;v<z;v++)w[v]="\n";v--;for(i=0;i<26;i++){for(j=0;j<z;j++)w[j]+=j-v?i-9?" ":"          ":y[i];~~(i/(z-1))%2?v++:v--;}if(z==1)w=y;console.log(w.join(""))

Here is a link to the jsfiddle, where you can test it.
edit: Made a console log for a monospaced font and removed the commas if input==1.

Answer 12

Perl 119

$s.=/26/?++$r&&$/:$m-1&&$r-abs$_%(2*$m-2)+1-$m?$"x(1+3*/15/):/15/?Perl:(A..Z)[$_]for(0..26)x($m=<>);print$s=~s/\s+$//rg

This program takes input from STDIN and prints the result to STDOUT.

And bonus - a version that is against the rules as it prints trailing spaces, but adds some interaction:

#!/usr/bin/perl -p
$r=0;$_=join'',map{/26/?++$r&&$/:$m-1&&$r-abs$_%(2*$m-2)+1-$m?$"x(1+3*/15/):/15/?Perl:(A..Z)[$_]}(0..26)x($m=$_)

...and some tests below:

1
ABCDEFGHIJKLMNOPerlQRSTUVWXYZ
2
 B D F H J L N Perl R T V X Z
A C E G I K M O    Q S U W Y
4
   D     J     Perl     V
  C E   I K   O    Q   U W
 B   F H   L N      R T   X Z
A     G     M        S     Y
6
     F         Perl         Z
    E G       O    Q       Y
   D   H     N      R     X
  C     I   M        S   W
 B       J L          T V
A         K            U

Answer 13

MetaPost (207)

numeric h,d,y;h=5;d=-1;y=1;u=5mm;pair p;string s;for x=1upto26:p:=(x,y)*u;s:=char(x+64);if x=13:s:="MetaPost";elseif x>13:p:=(x+2,y)*u;fi;label.rt(s,p);if(y=h)or(y=1):d:=d*-1;fi;if h=1:d:=0;fi;y:=y+d;endfor;

You can try it here.

Answer 14

Bash (213) (223)

read i
p=0
d=1
printf '\n%.0s' `seq $i`
for l in {A..Z}
do (($p==$i-1))&&((p=$p*-1))
(($i==1))&&p=0
(($p!=0))&&echo -en "\033[s\033[${p#-}A"
[[ "$l" == "B" ]]&&l="Bash"
echo -en $l"\033[s\033[${p#-}B"
((p++))
done

Minor tweaks and we're down to 213.

The original one had a minor bug. Tested and confirmed to work on bash version 4.2.37.

Thanks to @manatwork for pointing out the bug and some tips.

Answer 15

Haskell – 432 Bytes (unfortunately...)

This turned out to be much harder than I expected to accomplish purely, hence the hefty byte count. I am sure I (or someone) could do better, but I've spent too much time on this as it is. The golfed version is as follows:

import Data.Matrix
import Data.Vector(toList)
r=repeat
s h=y h$l(take 26(case h of 1->r 1;_->concat$r$[h,h-1..1]++[2..h-1]))['A'..'Z']$matrix h 32(const ' ')<|>matrix h 1(const '\n')
l _[]m=m
l(x:i)(r:e)m=l i e$u m(let z=26-length i in if z<9 then z else z+6)x r
l _ _ m=m
u m c r h=case h of 'H'->t c r"Haskell"m;_->setElem h(r,c)m
t _ _[]m=m
t c r(x:i)m=t(c+1)r i$setElem x(r,c)m
y h m=concat[toList$getRow x m|x<-[1..h]]

To run, load the code into ghci and execute putStr $ s Int where Int is your desired height. You could also add

import System.Environment
main :: IO ()
main = fmap (s . read . head) getArgs >>= putStr

under the imports in a text file, compile it with ghc, and pass the height as a command line argument. Ungolfed version:

import System.Environment
import Data.Matrix
import Data.Vector (toList)

main :: IO ()
main = fmap (sawtooth . read . head) getArgs >>= putStr

type M = Matrix Char

sawtooth :: Int -> String
sawtooth height = let mat     = matrix height 32 (const ' ') <|> matrix height 1 (const '\n')
                      numbers = take 26 (case height of 1 -> repeat 1
                                                        _ -> concat $ repeat $ [height,height-1..1]++[2..height-1])
                      chars   = ['A'..'Z']
                  in toString height $ makeMatrix numbers chars mat

makeMatrix :: [Int] -> String -> M -> M
makeMatrix []     []     mat = mat
makeMatrix (x:xs) (s:ss) mat = makeMatrix xs ss $ updateMatrix
                                                    mat (let z = 26 - length xs in if z < 9 then z else z + 6) x s
makeMatrix _      _      mat = mat

updateMatrix :: M -> Int -> Int -> Char -> M
updateMatrix mat col row char = case char of 'H' -> insertHaskell col row "Haskell" mat
                                             _   -> setElem char (row, col) mat

insertHaskell :: Int -> Int -> String -> M -> M
insertHaskell _   _   []     mat = mat
insertHaskell col row (x:xs) mat = insertHaskell (col+1) row xs $ setElem x (row, col) mat

toString ::Int -> M -> String
toString height mat = concat [ toList $ getRow x mat | x <- [1..height] ]

Answer 16

C#/LINQ:

using System;
using System.Linq;

namespace SawtoothAlphabet
{
    class Program
    {
        static void Main(string[] args)
        {
            int N = Int32.Parse(args[0]);   // eg 4
            Console.WriteLine(
                String.Join(Environment.NewLine, Enumerable.Range(0, N).Select(line =>
                    String.Concat(Enumerable.Range(0, 26).Select(i =>
                        line == Math.Abs(i % (2 * (N - 1)) - (N - 1))
                            ? (i == 2) ? "C#" : Char.ConvertFromUtf32(i + 'A')
                            : (i == 2) ? "  " : " ")
            ).TrimEnd())));
        }
    }
}

Brief explanation: Enumerable.Range(0, N).Select(...) causes a string to be generated for each line which is eventually concatenated into a single string with String.Join(Environment.NewLine, ...). For each line we cycle through all 26 characters with Enumerable.Range(0, 26).Select(...), the test at the start of the lambda expression determines whether to generate a character or space while the i==2 checks for "C" and converts it to either "C#" or two spaces depending on the line. The String.Concat(...) converts the resulting IEnumerable<char> for each line into a string before passing it to TrimEnd(...) to cleanly strip away any trailing whitespace.

Answer 17

K, 60 bytes

`0:'{+{x!y}.'+(26#{(!x),|1_!x-1}x;(((x-1)#" "),)'`c$65+!26)}

Pretty straightforward, and it looks like I just edged out the J solution. :)

First, generate an alphabet:

  `c$65+!26
"ABCDEFGHIJKLMNOPQRSTUVWXYZ"

And a sawtooth wave of an appropriate length:

  26#{(!x),|1_!x-1}5
0 1 2 3 4 3 2 1 0 1 2 3 4 3 2 1 0 1 2 3 4 3 2 1 0 1

Pad each letter of the alphabet with leading spaces:

  (((5-1)#" "),)'"ABC"
("    A"
 "    B"
 "    C")

Zip the alphabet and the square wave together, and rotate each row:

  +(0 1 2;("  A";"  B";"  C"))
((0
  "  A")
 (1
  "  B")
 (2
  "  C"))

  {x!y}.'+(0 1 2;("  A";"  B";"  C"))
("  A"
 " B "
 "C  ")

And the transposition of that is our answer:

  +{x!y}.'+(0 1 2;("  A";"  B";"  C"))
("  C"
 " B "
 "A  ")

Try it here in oK.

Answer 18

Javascript (204 185 150 145)

h=m=4;d=h>1?-1:0;o=[];for(a=0;a<35;a++){for(r=h;r;r--)o[r]=(o[r]||"")+(r==m?"ABCDEFGHIJavascriptKLMNOPQRSTUVWXYZ"[a]:" ");if(a<9||a>17){m+=d;if(m<2||m==h)d=-d}}console.log(o.join("\n"))

EDIT

Saved 13 bytes by not building an array and .join("\n"). Required flipping the for loops. Then, with help from C coding son, made code entirely too clever to save another 12 bytes.

Here is the readable version showing the logic change.

for (row = height; row; row--) {
    rowOfNextActiveLetter = 1;
    direction = height > 1 ? -1 : 0;
    output = "";
    for (a = 0; a < 35; a++) {
        output += (row == rowOfNextActiveLetter ? "ABCDEFGHIJavascriptKLMNOPQRSTUVWXYZ"[a] : "");
        if (a < 9 || a > 17) {
            rowOfNextActiveLetter -= direction;
            if (rowOfNextActiveLetter < 2 || rowOfNextActiveLetter == height)direction = -direction
        }
    }
    console.log(output)
}

Golfed (161):

for(r=h;r;r--){m=1;d=h>1?-1:0;o="";for(a=0;a<35;a++){o+=(r==m?"ABCDEFGHIJavascriptKLMNOPQRSTUVWXYZ"[a]:" ");if(a<9||a>17){m-=d;if(m<2||m==h)d=-d}}console.log(o)}

Golfed and obfuscated (149):

for(r=h;r;r--,console.log(o))for(m=1,d=h>1?1:0,o="",a=0;a<35;)if(o+=r==m?"ABCDEFGHIJavascriptKLMNOPQRSTUVWXYZ"[a]:" ",a<9|a++>17)d=(m+=d)<2|m==h?-d:d

Reversed Alphabet and loop (145):

for(r=h;r;r--,console.log(o))for(m=d=1,o="",a=35;a;)if(o+=" ZYXWVUTSRQPONMLKtpircsavaJIHGFEDCBA"[r==m?a:0],a<18|a-->26)d=h<2?0:(m+=d)<2|m==h?-d:d

Answer 19

PHP (216) (205)

New version:

$b=array_fill(65,26,' ');$b[80]='   ';$b=array_fill(0,$i,$b);for($j=--$i,$x=64;++$x<91;$i||$j=0){$b[$j][$x]=chr($x);$x==80&&$b[$j][$x]='PHP';$j+=($d=$j<($d?$i:1))*2-1;}echo join("\n",array_map('join',$b));

Old version:

$b=array_fill(0,$i,array_fill(0,28,' '));for($j=--$i,$x=-1;++$x<28;$i||$j=0){$q=&$b[$j];$q[$x]=chr($x-($x&16)/8+65);$x==15&&($q[++$x]='H')*$q[++$x]='P';$j+=($d=$j<($d?$i:1))*2-1;}echo join("\n",array_map('join',$b));

expects the variable $i to be the height.

Answer 20

C, 214 169 bytes, no trailing spaces

Thank to @edc65 and @tolos for their helpful suggestions.

#define C ((c-i+n-65)%z&&(c+i+n-67)%z)
n,i,m,c,z;main(){scanf("%d",&n);z=n<2?1:2*n-2;for(;i++<n;){for(m=c=64;c++<90;)m=C?m:c;for(c=64;c++<m;)putchar(C?32:c);puts("");}}

Answer 21

Perl 5 + -p, 79 bytes

Not the shortest, and I think there's a chance this isn't valid (as it uses ANSI escape sequences to position the letters...) - happy to remove if this is deemed invalid.

//;$_=$/x$_;$|-=++$-<$'?0:($-=1),$\.=$_.($|?".":".[A")x($'>1)for A..O,Perl,Q..Z

Try it online!

Explanation

The initial // is purely to store the current value of $_ in $' so $_ can be used in the body of the loop, saving a variable assignment. Then $_ is set to $/ (newline) $_ times, to move the cursor down the screen.

The main body of the program is the for loop that iterates over the alphabet, replacing P with Perl as a bareword. $| can be used as a binary toggle when decremented so if the post-incremented $- is less than $', subtract nothing, keeping the value the same, otherwise, set $- to 1 and decrement that. Next, append to $\ (which is automatically output after $_ at the end of the program thanks to -p) the current letter ($_ - implicitly set by the for loop) with if $| is truthy, form feed (to move the cursor down a line) otherwise \x1b[A to move the cursor up one line, $'>`` times. This last condition is to satisfy the requirement of input 1` being a straight line and feels the most out of place in the answer...

Answer 22

C: 142 139 characters

Dreadfully long, I expect to be able to shorten it a bit:

char a[26][27],i,j,p,d;main(n){for(scanf("%d",&n);i<26;d^=!p^p+2>n,p+=d-(d^n>1),i++)for(j=n;j--;a[j][i]=p^j?32:i+65);while(n)puts(a[--n]);}

Slightly more readable:

char a[26][27], i, j, p, d;
main(n) {
    for (
        scanf("%d", &n);
        i < 26;
        d ^= !p ^ p + 2 > n,
        p += d - (d ^ n > 1),
        i++
    )
        for (
            j = n;
            j--;
            a[j][i] = p ^ j
                ? 32
                : i + 65
        );

    while (n)
        puts(a[--n]);
}

Edit: I missed the "no trailing spaces"-rule, but I will be back.

Answer 23

Scala, 246 bytes

def f(h:Int)={val a=Seq.fill(h)(Array.fill(32)(' '));var(r,x,d)=(h-1,0,if(h==1)0 else-1);def p(c:Char){a(r)(x)=c;x+=1};for(c<-'A'to'Z'){p(c);if(c==83)"cala"foreach p;r+=d;if(r==0|r==h-1)d*= -1};for(z<-a)println(new String(z)replaceAll(" +$",""))}

reformatted and commented:

def f(h: Int) = {
  val a = Seq.fill(h)(Array.fill(32)(' '));
  // r - row; x - x coordinate, column; d - direction
  var (r, x, d) = (h - 1, 0, if(h==1) 0 else -1); 
  def p(c: Char) { // p for "put the character"
    a(r)(x) = c;
    x += 1
  };
  for(c <- 'A' to 'Z') { 
    p(c);
    if(c == 83)      // 83 == 'S'
      "cala" foreach p;
    r += d;
    if(r == 0 | r == h - 1)     // | is shorter than || 
      d *= -1
  };
  for(z <- a)
    println(new String(z) replaceAll (" +$", ""))  // trimming trailing whitespace
}

Results:

scala> f(4)
   D     J     P         V
  C E   I K   O Q       U W
 B   F H   L N   R     T   X Z
A     G     M     Scala     Y

scala> f(5)
    E       M           U
   D F     L N         T V
  C   G   K   O   Scala   W
 B     H J     P R         X Z
A       I       Q           Y

scala> f(1)
ABCDEFGHIJKLMNOPQRScalaTUVWXYZ

Answer 24

Python - 137

Input to be stored in i. e.g. i=8

l=[""]*h;k=j=0;y=h-1;exec'l[y]+=" "*(j-len(l[y]))+chr(k+65)+"ython"*(k==15);j=len(l[y]);y-=1^k/(h-(h>2))%2*-2;k+=1;'*26;print"\n".join(l)

Answer 25

Racket

Here's a clean functional version: suggestions to shorten it welcome.

(define (sawtooth n)
  (define (ST i d m lns)
    (if (null? m) 
        lns
        (let* ([v (first m)]
               [spc (make-string (string-length v) #\space)]
               [I (+ i d)])
          (ST I
              (if (or (zero? I) (= (sub1 n) I)) (* d -1) d)
              (rest m)
              (for/list ([ln lns] [j n])
                       (~a ln (if (= i j) v spc)))))))
  (displayln
   (string-join 
    (ST (sub1 n)
        (if (= n 1) 0 -1) 
        (string-split "A B C D E F G H I J K L M N O P Q Racket S T U V W X Y Z")
        (make-list n "\n")))))

Output

> (sawtooth 1)
ABCDEFGHIJKLMNOPQRacketSTUVWXYZ

> (sawtooth 2)
 B D F H J L N P Racket T V X Z 
A C E G I K M O Q      S U W Y 

> (sawtooth 8)
       H                  V     
      G I                U W    
     F   J              T   X   
    E     K            S     Y  
   D       L     Racket       Z 
  C         M   Q               
 B           N P                
A             O  

Answer 26

JavaScript, 168 bytes

f=(n,o=0,a=[...Array(n)])=>o<26?f(n,o+1,a.map((s='',i)=>s+(o-9?String.fromCharCode(65+o):'JavaScript').replace(i*i-(o%(2*n-2)-n+1)**2&&/./g,' '))):a.map(s=>s.trimEnd())

Try it online!

Returns an array of lines.

Ungolfed and explained:

f = (
  n,                 // input
  o = 0,             // offset for recursion (current step)
  a = [...Array(n)]  // array containing output lines
) =>
  o < 26 ?    // if not at end yet
    f(        // call function again
      n,      // with same input
      o + 1,  // increment offset
      a.map((s = '', i) =>  // add next letter to each line of a
        s +                 // current line
        (
          o - 9 ?                          // if letter is not J
            String.fromCharCode(65 + o) :  // current letter
            'JavaScript'                   // else 'JavaScript'
        ).replace(                          
          // replace with spaces if not at a triangular wave point
          i != Math.abs(o % (2 * n - 2) - n + 1) &&
          /./g, ' '
        )
      )
    ) :
    a.map(s => s.trimEnd())  // return array with trailing spaces removed

Answer 27

Scala, 332 bytes

Golfed version. Try it online!

object P extends App{val h=args(0).toInt;val x="ABCDEFGHIJKLMNOPQRSTUVWXYZ";val l=Array.fill[String](h)("");var i=h-1;var d= -1;for(c<-x.toCharArray){for(n<-l.indices){val k=if(n==i){if(c=='S')"Scala"else c.toString}else{if(c=='S')"    "else" "};l(n)+=k};if(i==0&&d== -1)d=1;if(i==h-1&&d==1)d= -1;if(h>1)i+=d};for(s<-l){println(s)}}

Ungolfed version. Try it online!

object P extends App {
  val h = args(0).toInt
  val x = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
  val l = Array.fill[String](h)("")
  var i = h - 1
  var d = -1
  for (c <- x.toCharArray) {
    for (n <- l.indices) {
      val k = if (n == i) {
        if (c == 'S') "Scala" else c.toString
      } else {
        if (c == 'S') "    " else " "
      }
      l(n) += k
    }
    if (i == 0 && d == -1) d = 1
    if (i == h - 1 && d == 1) d = -1
    if (h > 1) i += d
  }
  for (s <- l) {
    println(s)
  }
}