First Last Last First

Question

Challenge

The task is simple. Given an array and a first and last value: Return the first of the last after the first, and the last of the first before the last.

---

Or simply: Given an array, var1, var2.

Example Array:

[ var2, , var1, , var2, , var2, var1, var2, ]

Return:

• The index of the first var2 on the right side of the first var1 that appears in the array.

[ var2, , first var1, , first var2, , second var2, var1, third var2, ]

• The index of the first var1 on the left side of the last var2 that appears in the array.

[ var2, , second var1, , var2, , var2, first var1,last var2, ]

Input

Two distinct positive integers

Array of positive integers

Output

Index of answers, in order

Rules

The array will contain at least one of each variable (minimum size of 2)

Assume inputs work

Example: 0, 1 [1, 0] or similar would fail

IO is flexible

Examples

Input
First = 2; Last = 4; [0, 2, 4, 2, 3, 1, 4, 0, 1, 2, 4, 9]

Output
2, 9

---

Input
First = 4; Last = 2; [0, 2, 4, 2, 3, 1, 4, 0, 1, 2, 4, 9]

Output
3, 6

---

Input
First = 0; Last = 1; [0, 1]

Output
1, 0

Answer 1

Python 2, 72 bytes

def f(x,y,a):i=a.index;j=a[::-1].index;print i(y,i(x)),len(a)+~j(x,j(y))

Try it online!

Answer 2

APL (Dyalog Classic), 29 27 bytes

⊃{(⊃⍵~⍳⊃⍺),⊃⌽⍺∩⍳⊃⌽⍵}/⍸¨⎕=⊂⎕

Try it online!

prompts for the array and then for var1,var2

Answer 3

JavaScript (ES6), 63 bytes

(x,y,a)=>a.map(P=(v,i)=>v-y?v-x?0:a=i:1/(p=a)?P=+P||i:0)&&[P,p]

Try it online!

Commented

(x, y, a) =>          // given the two integers x, y and the array a[]
  a.map(P =           // initialize P to a non-numeric value
            (v, i) => // for each value v at position i in a[]:
    v - y ?           //   if v is not equal to y:
      v - x ?         //     if v is not equal to x:
        0             //       do nothing
      :               //     else (v = x):
        a = i         //       save the current position in a
    :                 //   else (v = y):
      1 / (p = a) ?   //     update p to a (last position of x); if p is numeric (>= 0):
        P = +P || i   //       unless P is also already numeric, update it to i
                      //       (if P is numeric, it's necessarily greater than 0 because
                      //       we've also seen x before; that's why +P works)
      :               //     else:
        0             //       do nothing
  )                   // end of map()
  && [P, p]           // return [P, p]

Alternate versions

Using JS built-ins, a more straightforward answer is 79 bytes:

(x,y,a)=>[a.indexOf(y,a.indexOf(x)),a.slice(0,a.lastIndexOf(y)).lastIndexOf(x)]

which can be slightly compressed to 75 bytes:

(x,y,a)=>[a.indexOf(y,a.indexOf(x)),a.slice(0,a[L='lastIndexOf'](y))[L](x)]

Try it online!

Edit: @Neil managed to reduce it to a very nice 67-byte:

(x,y,a,f=s=>a[z=y,y=x,x=z,s+=`ndexOf`](x,a[s](y)))=>[f`i`,f`lastI`]

Try it online!

Answer 4

Jelly, 17 bytes

ẹⱮṚ>Ƈ<Ƈƭ"1,0ị"$⁺Ʋ

Try it online!

Answer 5

Python 3, 97 93 bytes

-4 bytes thanks to ovs

def h(f,l,a,I=list.index):j=I(a,f);i=len(a)+~I(a[::-1],l);print(I(a[j:],l)+j,i-I(a[i::-1],f))

Try it online!

Answer 6

Japt, 27 25 24 bytes

Inspired in @Arnauld answer

Thanks @Shaggy -2 bytes and @ETHproductions -1 byte

I just started with japt so it must be a better way.\

[WsX=WbU)bV +XWsTWaV)aU]

Try it online!

Answer 7

APL (Dyalog Unicode), 42 bytes^SBCS

Anonymous tacit infix function. Takes var1,var2 as left argument and the array as right argument.

{⍸<\(⍵=⊃⌽⍺)∧∨\⍵=⊃⍺},{⍸⌽<\(⍵=⊃⍺)∧∨\⍵=⊃⌽⍺}∘⌽

Try it online!

Answer 8

R, 81 bytes

function(a,b,v,x=which(v==b),y=which(v==a))c(x[x>y[1]][1],tail(y[y<tail(x,1)],1))

Try it online!

(1-indexed)

Answer 9

MATL, 27 bytes

y=Y>/ti=PY>P/t3G=f1)w2G=f0)

Try it online!

Alternately for the same bytecount:

27 bytes

y=Y>yi=*f1)y3G=PY>Pb2G=*f0)

Try it online!

The second one is easier to explain:

y   % implicitly get the first two inputs (the array and var1),
    %  and duplicate the first input
    %  stack: [[0 2 4 2 3 1 4 0 1 2 4 9] 2 [0 2 4 2 3 1 4 0 1 2 4 9]]
=   % compare and return logical (boolean) array
    %  stack: [[0 2 4 2 3 1 4 0 1 2 4 9] [0 1 0 1 0 0 0 0 0 1 0 0]]
Y>  % cumulative maximum - make all values after the first 1 also 1s
    %  stack: [[0 2 4 2 3 1 4 0 1 2 4 9] [0 1 1 1 1 1 1 1 1 1 1 1]]
    %  now we have 1s in positions at and after the first time var1 appears
y   % duplicate 2nd element in stack
    %  stack: [[0 2 4 2 3 1 4 0 1 2 4 9] [0 1 1 1 1 1 1 1 1 1 1 1] [0 2 4 2 3 1 4 0 1 2 4 9]]
i=  % compare with the next input (var2), returning a boolean array
    % stack: [[0 2 4 2 3 1 4 0 1 2 4 9] [0 1 1 1 1 1 1 1 1 1 1 1] [0 0 1 0 0 0 1 0 0 0 1 0]]
*   % multiply the two boolean arrays - so we'll have 1s only where var2 was present after the first occurrence of var1
    % stack: [[0 2 4 2 3 1 4 0 1 2 4 9] [0 0 1 0 0 0 1 0 0 0 1 0]]
f1) % find the index of the first 1 in that (this is our first result value)

The second part of the code does the same thing, except for these changes:

• use 2G for second input (var1) and 3G first 3rd input (var2) instead of implicit input or i, since those have been consumed
• use PY>P (flip array left-to-right, get cumulative maximum, flip back) instead of Y>, to get 1s before the last occurrence instead of after the first occurrence
• use f0) to get last place where both conditions are true, instead of first place (works because MATL uses modular indexing, so 0 is taken to refer to the last index of the array)

Answer 10

MATLAB (80 bytes)

Input is x, y, and a. Since MATLAB is 1-indexed, you should add 1 to test cases.

xi=find(a==x);
yi=find(a==y);
yi(find(yi>xi(1),1))
xi(find(xi<yi(end),1,'last'))

Test case:

x=4
y=2
a =  [0, 2, 4, 2, 3, 1, 4, 0, 1, 2, 4, 9]

% 
xi=find(a==x);
yi=find(a==y);
yi(find(yi>xi(1),1))
xi(find(xi<yi(end),1,'last'))

ans =

     4


ans =

     7

Answer 11

Java 8, 114 bytes

A lambda taking a java.util.List<Integer> and two ints (var1, var2) and returning a comma-separated pair.

(a,f,l)->a.indexOf(f)+a.subList(a.indexOf(f),a.size()).indexOf(l)+","+a.subList(0,a.lastIndexOf(l)).lastIndexOf(f)

Try It Online

Answer 12

Kotlin, 132 bytes

{f:Int,l:Int,a:Array<Int>->{var p=a.indexOfFirst{it==f}
while(a[p]!=l)p++
var i=a.indexOfLast{it==l}
while(a[i]!=f)i--
Pair(p,i)}()}

Try it online!

Answer 13

Julia, 71 64 bytes

thanks to sundar and his find(A.==x)[] instead of findfirst(A,x)).

.

(A,x,y)->findnext(A,y,find(A.==x)[]),findprev(A,x,findlast(A,y))