Cooking with Code

Haskell, 48 bytes

foldl(?) is an anonymous function taking two list arguments and returning a list, with all elements of the same (Eq-comparable) type.

Use as foldl(?)["pepper", "paprika", "salt", "cumin", "oregano"]["oregano", "cumin", "cumin", "salt", "salt", "salt"].

foldl(?)
(x:y:r)?n|y==n=y:x:r|s<-y:r=x:s?n
s?n=s

Try it online!

• foldl(?) s m starts with the (spice rack) list s and combines it with each element (spice) from m in order, using the operator ?.
• s?n uses the spice n from the spice rack s and returns the resulting spice rack.
  • If s has at least two elements, ? checks whether the second one is equal to n, and if so switches the first two elements. If not equal, ? keeps the first element fixed and recurses on the rest.
  • If s has at most one element, ? returns it unchanged.

Chef, 875 843 bytes

S.

Ingredients.
1 g T
0 g Z
J
I

Method.
Put Z into mixing bowl.Take I from refrigerator.B I.Put I into mixing bowl.Take I from refrigerator.B until bed.Take I from refrigerator.V I.Put Z into 2nd mixing bowl.N T.Fold J into mixing bowl.Put J into mixing bowl.Remove I.Fold T into mixing bowl.Put J into 2nd mixing bowl.N until ned.Fold T into mixing bowl.Put T into mixing bowl.Fold J into 2nd mixing bowl.Put J into mixing bowl.C T.Stir for 1 minute.Put Z into mixing bowl.Fold T into mixing bowl.C until ced.Fold T into 2nd mixing bowl.G T.Put T into mixing bowl.Fold T into 2nd mixing bowl.G until ged.Put I into mixing bowl.Fold T into mixing bowl.Take I from refrigerator.V until ved.Fold T into mixing bowl.L T.Put T into 2nd mixing bowl.Fold T into mixing bowl.L until led.Pour contents of 2nd mixing bowl into baking dish.

Serves 1.

-32 bytes thanks to Jonathan Allan by removing the where I wouldn't think it will work.

Chef has no string types, so the ingredients are positive integers. 0 is used to separate the starting list from used ingredients and to end the used ingredients list. See the TIO link for an example.

Pseudocode explaination:

A, B: stack
T, J, IN: int
T = 1
A.push(0) // used as the marker for the bottom of the stack
IN = input() // input the first list
while(IN):
    A.push(IN)
    IN = input()
IN = input() // for each used ingredient
while(IN):
    B.push(0)
    while(T): // move all ingredients up to and including the one we are moving right now to the second stack
        T = A.peek() - IN
        B.push(A.pop())
    A.push(B.pop())
    if(A.peekUnderTop() != 0):
        A.swapTopTwoItems()
    T = B.pop() // move the ingredients from the second stack back to the first
    while(T):
        A.push(T)
        T = B.pop()
    T = IN // to make it non-zero for next iteration
    IN = input(0
print(A.inverted())

Try it online!

JavaScript, 61 bytes

a=>b=>b.map(v=>(p=a.indexOf(v))&&a.splice(p-1,2,a[p],a[p-1]))

Input format:

• f(list_of_spices)(list_of_what_spices_got_used)
• two list are array of string

Output:

• list_of_spices is modified in-place.

f=
a=>b=>b.map(v=>(p=a.indexOf(v))&&a.splice(p-1,2,a[p],a[p-1]))

a = ['pepper', 'paprika', 'salt', 'cumin', 'oregano'];
b = ['salt', 'salt', 'salt', 'salt', 'salt', 'salt', 'salt', 'salt', 'salt', 'salt', 'salt', 'salt', 'oregano'];
f(a)(b);
document.write(JSON.stringify(a))