Diberikan string Sdan daftar indeks X, modifikasi Sdengan menghapus elemen pada setiap indeks Ssaat menggunakan hasil itu sebagai nilai baru S.

Misalnya, diberikan S = 'codegolf'dan X = [1, 4, 4, 0, 2],

0 1 2 3 4 5 6 7  |
c o d e g o l f  |  Remove 1
c d e g o l f    |  Remove 4
c d e g l f      |  Remove 4
c d e g f        |  Remove 0
d e g f          |  Remove 2
d e f

Tugas Anda adalah melakukan proses ini, mengumpulkan nilai Ssetelah setiap operasi, dan menampilkannya masing-masing pada baris baru secara berurutan. Jawaban akhirnya adalah

S = 'codegolf'
X = [1, 4, 4, 0, 2]

Answer:

codegolf
cdegolf
cdeglf
cdegf
degf
def

• Ini kode-golf jadi buat kode Anda sesingkat mungkin.
• Anda dapat mengasumsikan bahwa nilai-nilai dalam Xselalu merupakan indeks yang valid S, dan Anda dapat menggunakan pengindeksan berbasis 0 atau 1.
• String hanya akan berisi [A-Za-z0-9]
• Entah Satau xmungkin dengan kosong. Jika Skosong, maka itu xjuga harus kosong.
• Anda juga dapat mengambil Ssebagai daftar karakter, bukan string.
• Anda dapat mencetak output atau mengembalikan daftar string. Ruang putih terkemuka dan tertinggal dapat diterima. Segala bentuk output baik-baik saja asalkan mudah dibaca.

Uji Kasus

S = 'abc', x = [0]
'abc'
'bc'

S = 'abc', x = []
'abc'

S = 'abc', x = [2, 0, 0]
'abc'
'ab'
'b'
''

S = '', x = []
''

S = 'codegolfing', x = [10, 9, 8, 3, 2, 1, 0]
'codegolfing'
'codegolfin'
'codegolfi'
'codegolf'
'codgolf'
'cogolf'
'cgolf'
'golf'

Haskell, 38 33 byte

s#i=take i s++drop(i+1)s
scanl(#)

Lurus ke depan: berulang kali ambil elemen sebelum dan sesudah indeks i, bergabung kembali dan kumpulkan hasilnya.

Cobalah online!

Sunting: @ Lynn menyimpan 5 byte. Terima kasih!

JavaScript (ES6), 57 50 48 45 42 byte

Mengambil string sebagai larik karakter individu, menghasilkan larik yang berisi string yang dipisahkan koma dari aslinya diikuti oleh subarray string yang dipisahkan koma untuk setiap langkah.

s=>a=>[s+"",a.map(x=>s.splice(x,1)&&s+"")]

• 3 byte yang disimpan berkat Arnauld yang menyarankan saya menyalahgunakan spesifikasi output yang longgar lebih dari yang sudah saya lakukan sebelumnya, yang membuat saya lebih menyalahgunakannya untuk penghematan 3 byte lainnya.

Menguji

o.innerText=JSON.stringify((f=

s=>a=>[s+"",a.map(x=>s.splice(x,1)&&s+"")]

)([...i.value="codegolf"])(j.value=[1,4,4,0,2]));oninput=_=>o.innerText=JSON.stringify(f([...i.value])(j.value.split`,`))

label,input{font-family:sans-serif;font-size:14px;height:20px;line-height:20px;vertical-align:middle}input{margin:0 5px 0 0;width:100px;}

<label for=i>String: </label><input id=i><label for=j>Indices: </label><input id=j><pre id=o>

Penjelasan

Kami mengambil dua input melalui parameter s(array string) dan a(array integer) dalam sintaks currying, artinya kita memanggil fungsi denganf(s)(a) .

Kami membangun array baru dan memulainya dengan yang asli s. Namun, sebagaisplice metode yang akan kita gunakan nanti memodifikasi array, kita perlu membuat salinannya, yang bisa kita lakukan dengan mengubahnya menjadi string (cukup tambahkan string kosong).

Untuk menghasilkan subarray, kita mapmelewati array integer a(di mana xinteger saat ini) dan, untuk setiap elemen, kita splice1 elemen dari s, dimulai dari indeks x. Kami mengembalikan yang dimodifikasi s, kembali membuat salinannya dengan mengubahnya menjadi string.

Japt , 6 byte

åjV uV

Uji secara online!

Penjelasan

UåjV uV   Implicit: U = array of integers, V = string
Uå        Cumulatively reduce U by
  j         removing the item at that index in the previous value,
   V        with an initial value of V.
     uV   Push V to the beginning of the result.

Kalau tidak:

uQ åjV

UuQ       Push a quotation mark to the beginning of U.
    å     Cumulatively reduce by
     j      removing the item at that index in the previous value,
      V     with an initial value of V.

Ini berfungsi karena menghapus item pada indeks "tidak melakukan apa-apa dan mengembalikan string asli.

Sekam , 7 byte

G§+oh↑↓

Mengambil string terlebih dahulu, lalu indeks (berbasis 1). Cobalah online!

Penjelasan

G§+oh↑↓
G        Scan from left by function:
           Arguments are string, say s = "abcde", and index, say i = 3.
      ↓    Drop i elements: "de"
     ↑     Take i elements
   oh      and drop the last one: "ab"
 §+        Concatenate: "abde"
         Implicitly print list of strings on separate lines.

Python 2 , 43 byte

s,i=input()
for i in i+[0]:print s;s.pop(i)

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Segala bentuk output baik-baik saja asalkan mudah dibaca.

Jadi ini dicetak sebagai daftar karakter.

Python 2 , 47 byte

Ini bisa disingkat menjadi 43 byte , seperti yang ditunjukkan oleh @LuisMendo, tapi itu sudah solusi dari @ ErktheOutgolfer.

a,b=input();print a
for i in b:a.pop(i);print a

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Java 8, 78 byte

Ini adalah lambda kari, dari int[]ke konsumen StringBuilderatau StringBuffer. Output dicetak ke standar keluar.

l->s->{System.out.println(s);for(int i:l)System.out.println(s.delete(i,i+1));}

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05AB1E , 11 byte

v=ā0m0yǝÏ},

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v           # For each index:
 =          #   Print without popping
  ā         #   Push range(1, len(a) + 1)
   0m       #   Raise each to the power of 0. 
            #   This gives a list of equal length containing all 1s
     0yǝ    #   Put a 0 at the location that we want to remove
        Ï   #   Keep only the characters that correspond to a 1 in the new list
         }, # Print the last step

Mathematica, 70 byte

(s=#;For[t=1,t<=Length@#2,Print@s;s=StringDrop[s,{#2[[t]]+1}];t++];s)&

Cobalah online!

R , 46 32 byte

function(S,X)Reduce(`[`,-X,S,,T)

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Mengambil input sebagai daftar karakter dan Xberbasis 1. Reduceadalah setara dengan R fold, fungsi dalam hal ini [adalah subset. Berulang-ulang -Xkarena pengindeksan negatif dalam R menghilangkan elemen, dan initdiatur ke S, dengan accum=TRUEdemikian kami mengakumulasikan hasil antara.

R , 80 byte

function(S,X,g=substring)Reduce(function(s,i)paste0(g(s,0,i-1),g(s,i+1)),X,S,,T)

2-argument function, takes X 1-indexed. Takes S as a string.

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Haskell, 33 bytes

scanl(\s i->take i s++drop(i+1)s)

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PowerShell, 94 84 bytes

param($s,$x)$a=[Collections.Generic.list[char]]$s;$x|%{-join$a;,$a|% r*t $_};-join$a

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Takes input $s as a string and $x as an explicit array. We then create $a based on $s as a list.

Arrays in PowerShell are fixed size (for our purposes here), so we need to use the lengthy [System.Collections.Generic.list] type in order to get access to the .removeAt() function, which does exactly what it says on the tin.

I sacrificed 10 bytes to include two -join statements to make the output pretty. OP has stated that outputting a list of chars is fine, so I could output just $a rather than -join$a, but that's really ugly in my opinion.

Saved 10 bytes thanks to briantist.

Python 2, 50 bytes

• Takes in string and list of indices

s,i=input()
for j in i+[0]:print s;s=s[:j]+s[j+1:]

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05AB1E, 9 7 bytes

=svõyǝ=

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=s        # Print original string, swap with indices.
  v       # Loop through indices...
   õyǝ    # Replace current index with empty string.

-2 thanks to idea from @ETHProductions.

Retina, 58 bytes

¶\d+
¶¶1$&$*
+1`(?=.*¶¶.(.)*)(((?<-1>.)*).(.*)¶)¶.*
$2$3$4

Try it online! Explanation:

¶\d+

Match the indices (which are never on the first line, so are always preceded by a newline).

¶¶1$&$*

Double-space the indices, convert to unary, and add 1 (because zeros are hard in Retina).

+1`

Repeatedly change the first match, which is always the current value of the string.

   (?=.*¶¶.(.)*)

Retrieve the next index in $#1.

                (           .    ¶)

Capture the string, including the $#1th character and one newline.

                 ((?<-1>.)*) (.*)

Separately capture the prefix and suffix of the $#1th character of the string.

                                   ¶.*

Match the index.

$2$3$4

Replace the string with itself and the index with the prefix and suffix of the $#1th character.

Pyth, 8 bytes

.u.DNYEz

Demonstration

Reduce, starting with the string and iterating over the list of indices, on the deletion function.

PowerShell, 54 58 bytes

param($s,$x),-1+$x|%{$z=$_;$i=0;-join($s=$s|?{$z-ne$i++})}

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Explanation

Takes input as a char array ([char[]]).

Iterates through the array of indices ($x) plus an injected first element of -1, then for each one, assigns the current element to $z, initializes $i to 0, then iterates through the array of characters ($s), returning a new array of only the characters whose index ($i) does not equal (-ne) the current index to exclude ($z). This new array is assigned back to $s, while simultaneously being returned (this happens when the assignment is done in parentheses). That returned result is -joined to form a string which is sent out to the pipeline.

Injecting -1 at the beginning ensures that the original string will be printed, since it's the first element and an index will never match -1.

q/kdb+, 27 10 bytes

Solution:

{x _\0N,y}

Examples:

q){x _\0N,y}["codegolf";1 4 4 0 2]
"codegolf"
"cdegolf"
"cdeglf"
"cdegf"
"degf"
"def"
q){x _\0N,y}["abc";0]
"abc"
"bc"
q){x _\0N,y}["abc";()]
"abc"
q){x _\0N,y}["abc";2 0 0]
"abc"
"ab"
,"b"
""    
q){x _\0N,y}["";()]
""

Explanation:

Takes advantage of the converge functionality \ as well as drop _.

{x _\0N,y}
{        } / lambda function, x and y are implicit variables
     0N,y  / join null to the front of list (y), drop null does nothing
   _\      / drop over (until result converges) printing each stage
 x         / the string (need the space as x_ could be a variable name)

Notes:

If we didn't need to print the original result, this would be 2 bytes in q:

q)_\["codegolfing";10 9 8 3 2 1 0]
"codegolfin"
"codegolfi"
"codegolf"
"codgolf"
"cogolf"
"cgolf"
"golf"

Perl 5, 55 bytes (54 + "-l")

sub{print($s=shift);for(@_){substr$s,$_,1,"";print$s}}

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MATL, 8 bytes

ii"t[]@(

Indexing is 1-based.

Try it online! Or verify the test cases.

Explanation

i      % Input string. Input has to be done explicitly so that the string
       % will be displayed even if the row vector of indices is empty
i      % Input row vector of indices
"      % For each
  t    %   Duplicate current string
  []   %   Push empty array
  @    %   Push current index
  (    %   Assignment indexing: write [] to string at specified index
       % End (implicit). Display stack (implicit)

C# (.NET Core), 87 87 74 70 bytes

S=>I=>{for(int i=0;;S=S.Remove(I[i++],1))System.Console.WriteLine(S);}

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Just goes to show that recursion isn't always the best solution. This is actually shorter than my original invalid answer. Still prints to STDOUT rather than returning, which is necessary because it ends with an error.

-4 bytes thanks to TheLethalCoder

C (gcc), 99 bytes

j;f(s,a,i)char*s;int*a;{puts(s);for(j=0;j<i;j++){memmove(s+a[j],s+a[j]+1,strlen(s)-a[j]);puts(s);}}

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Takes the string, array, and the length of the array.

Pyth, 10 bytes

Rule changes saved me 1 byte:

V+E0Q .(QN

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Pyth, 11 bytes

V+E0sQ .(QN

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Gaia, 9 bytes

+⟪Seḥ+⟫⊣ṣ

^{I should really add a "delete at index" function...}

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Explanation

+          Add the string to the list
 ⟪Seḥ+⟫⊣   Cumulatively reduce by this block:
  S         Split around index n
   e        Dump the list
    ḥ       Remove the first char of the second part
     +      Concatenate back together
        ṣ  Join the result with newlines

V, 12 bytes

òdt,GÙ@-|xHx

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This is 1-indexed, input is like:

11,10,9,4,3,2,1,
codegolfing

Explanation

ò              ' <M-r>ecursively until error
 dt,           ' (d)elete (t)o the next , (errors when no more commas)
    G          ' (G)oto the last line
     Ù         ' Duplicate it down
        |      ' Goto column ...
      @-       ' (deleted number from the short register)
         x     ' And delete the character there
          H    ' Go back home
           x   ' And delete the comma that I missed

Swift 3, 80 bytes

func f(l:[String],c:[Int]){var t=l;for i in c{print(t);t.remove(at:i)};print(t)}

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Pyth, 8 bytes

+zm=z.Dz

Test suite!

explanation

+zm=z.DzdQ    # implicit: input and iteration variable
  m      Q    # for each of the elements of the first input (the array of numbers, Q)
     .Dzd     # remove that index from the second input (the string, z)
   =z         # Store that new value in z
+z            # prepend the starting value

Python 2, 54

X,S=input()
for a in X:
    print S
    S=S[:a]+S[a+1:]
print S

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APL, 31 30 28 bytes

{⎕←⍺⋄⍵≡⍬:⍬⋄⍺[(⍳⍴⍺)~⊃⍵]∇1↓⍵}

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C# (Mono), 85 bytes

s=>a=>{for(int i=-2;++i<a.Count;)System.Console.WriteLine(s=i<0?s:s.Remove(a[i],1));}

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