You are given a string
expression representing a Lisp-like expression to return the integer value of.
The syntax for these expressions is given as follows.
(let v1 e1 v2 e2 ... vn en expr), where
letis always the string
"let", then there are 1 or more pairs of alternating variables and expressions, meaning that the first variable
v1is assigned the value of the expression
e1, the second variable
v2is assigned the value of the expression
e2, and so on sequentially; and then the value of this let-expression is the value of the expression
(add e1 e2)where
addis always the string
"add", there are always two expressions
e1, e2, and this expression evaluates to the addition of the evaluation of
e1and the evaluation of
(mult e1 e2)where
multis always the string
"mult", there are always two expressions
e1, e2, and this expression evaluates to the multiplication of the evaluation of
e1and the evaluation of
Input: (add 1 2) Output: 3 Input: (mult 3 (add 2 3)) Output: 15 Input: (let x 2 (mult x 5)) Output: 10 Input: (let x 2 (mult x (let x 3 y 4 (add x y)))) Output: 14 Explanation: In the expression (add x y), when checking for the value of the variable x, we check from the innermost scope to the outermost in the context of the variable we are trying to evaluate. Since x = 3 is found first, the value of x is 3. Input: (let x 3 x 2 x) Output: 2 Explanation: Assignment in let statements is processed sequentially. Input: (let x 1 y 2 x (add x y) (add x y)) Output: 5 Explanation: The first (add x y) evaluates as 3, and is assigned to x. The second (add x y) evaluates as 3+2 = 5. Input: (let x 2 (add (let x 3 (let x 4 x)) x)) Output: 6 Explanation: Even though (let x 4 x) has a deeper scope, it is outside the context of the final x in the add-expression. That final x will equal 2. Input: (let a1 3 b2 (add a1 1) b2) Output 4 Explanation: Variable names can contain digits after the first character.
expressionis well formatted: There are no leading or trailing spaces, there is only a single space separating different components of the string, and no space between adjacent parentheses. The expression is guaranteed to be legal and evaluate to an integer.
expressionis at most 2000. (It is also non-empty, as that would not be a legal expression.)
Intuition and Algorithm\n
This question is relatively straightforward in terms of the idea of the solution, but presents substantial difficulties in the implementation.\n
Expressions may involve the evaluation of other expressions, which motivates a recursive approach.\n
One difficulty is managing the correct scope of the variables. We can use a stack of hashmaps. As we enter an inner scope defined by parentheses, we need to add that scope to our stack, and when we exit, we need to pop that scope off.\n
evaluate function will go through each case of what form the
expression could take.
If the expression starts with a digit or \'-\', it\'s an integer: return it.\n
If the expression starts with a letter, it\'s a variable. Recall it by checking the current scope in reverse order.\n
Otherwise, group the tokens (variables or expressions) within this expression by counting the "balance"
bal of the occurrences of
\'(\' minus the number of occurrences of
\')\'. When the balance is zero, we have ended a token. For example,
(add 1 (add 2 3)) should have tokens
\'(add 2 3)\'.
For add and mult expressions, evaluate each token and return the addition or multiplication of them.\n
For let expressions, evaluate each expression sequentially and assign it to the variable in the current scope, then return the evaluation of the final expression.\n
Time Complexity: , where is the length of
expression. Each expression is evaluated once, but within that evaluation we may search the entire scope.
Space Complexity: . We may pass new strings to our
evaluate function when making intermediate evaluations, each of length . With effort, we could reduce the total space complexity to with interning or passing pointers.
Analysis written by: @awice.\n