Algorithm
Problem Name: 736. Parse Lisp Expression
Problem Link: https://leetcode.com/problems/parse-lisp-expression/
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.
- An expression is either an integer, let expression, add expression, mult expression, or an assigned variable. Expressions always evaluate to a single integer.
- (An integer could be positive or negative.)
- A let expression takes the form
"(let v1 e1 v2 e2 ... vn en expr)", where let is always the string"let", then there are one or more pairs of alternating variables and expressions, meaning that the first variablev1is assigned the value of the expressione1, the second variablev2is assigned the value of the expressione2, and so on sequentially; and then the value of this let expression is the value of the expressionexpr. - An add expression takes the form
"(add e1 e2)"where add is always the string"add", there are always two expressionse1,e2and the result is the addition of the evaluation ofe1and the evaluation ofe2. - A mult expression takes the form
"(mult e1 e2)"where mult is always the string"mult", there are always two expressionse1,e2and the result is the multiplication of the evaluation of e1 and the evaluation of e2. - For this question, we will use a smaller subset of variable names. A variable starts with a lowercase letter, then zero or more lowercase letters or digits. Additionally, for your convenience, the names
"add","let", and"mult"are protected and will never be used as variable names. - Finally, there is the concept of scope. When an expression of a variable name is evaluated, within the context of that evaluation, the innermost scope (in terms of parentheses) is checked first for the value of that variable, and then outer scopes are checked sequentially. It is guaranteed that every expression is legal. Please see the examples for more details on the scope.
Example 1:
Input: expression = "(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.
Example 2:
Input: expression = "(let x 3 x 2 x)" Output: 2 Explanation: Assignment in let statements is processed sequentially.
Example 3:
Input: expression = "(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.
Constraints:
1 <= expression.length <= 2000- There are no leading or trailing spaces in
expression. - All tokens are separated by a single space in
expression. - The answer and all intermediate calculations of that answer are guaranteed to fit in a 32-bit integer.
- The expression is guaranteed to be legal and evaluate to an integer.
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#define TYPE_INT 0
#define TYPE_ID 1
typedef struct {
char *p;
int len;
} idn_t;
typedef struct {
int type; // 0: identifier, 1: value
union {
idn_t id;
int num;
} u;
} expr_t;
typedef struct sym_s {
idn_t id;
int val;
int scope;
struct sym_s *list;
} sym_t;
typedef struct {
char *input;
sym_t *sym;
} p_t;
#define IS_LET(E) ((E)[0] == 'l' && \
(E)[1] == 'e' && \
(E)[2] == 't' && \
(E)[3] == ' ')
#define IS_ADD(E) ((E)[0] == 'a' && \
(E)[1] == 'd' && \
(E)[2] == 'd' && \
(E)[3] == ' ')
#define IS_MULT(E) ((E)[0] == 'm' && \
(E)[1] == 'u' && \
(E)[2] == 'l' && \
(E)[3] == 't' && \
(E)[4] == ' ')
#define IS_NUM(E) ((E)[0] >= '0' && \
(E)[0] < = '9')
char *next_input(char *input) {
int n = 0;
char c;
while (c = *(input ++)) {
if (c == '(') n ++;
else if (c == ')') {
if (n) n --;
else {
*(input - 1) = 0;
break;
}
}
}
if (*input == ' ') input ++;
//assert(0);
return input;
}
idn_t parse_identifier(p_t *p) {
char c;
idn_t id;
id.p = p->input;
do {
c = *(++ p->input);
} while (c != 0 && c != ' ');
id.len = p->input - id.p;
if (c == ' ') p->input ++;
return id;
}
expr_t parse_num(p_t *p) {
expr_t expr;
int neg = 0;
if (*p->input == '-') {
neg = 1;
p->input ++;
}
expr.type = TYPE_INT;
expr.u.num = 0;
do {
expr.u.num = expr.u.num * 10 + *p->input - '0';
p->input ++;
} while (IS_NUM(p->input));
if (neg) {
expr.u.num = 0 - expr.u.num;
}
if (*p->input == ' ') p->input ++;
return expr;
}
expr_t resolve(p_t *p, expr_t a) {
sym_t *sym;
if (a.type == TYPE_ID) {
sym = p->sym;
while (a.u.id.len != sym->id.len ||
strncmp(a.u.id.p, sym->id.p, sym->id.len)) sym = sym->list;
a.type = TYPE_INT;
a.u.num = sym->val;
}
return a;
}
expr_t parse(p_t *, int);
expr_t parse_let(p_t *p, int d) {
char c;
sym_t *sym;
expr_t a, b;
while (c = *p->input) {
if (c == ' ') {
p->input ++;
continue;
}
a = parse(p, d);
if (a.type == TYPE_INT) break;
if (*p->input) {
b = parse(p, d);
b = resolve(p, b);
sym = malloc(sizeof(*sym));
//assert(sym);
sym->id = a.u.id;
sym->val = b.u.num;
sym->scope = d;
sym->list = p->sym;
p->sym = sym;
}
}
if (a.type == TYPE_ID) {
a = resolve(p, a);
}
sym = p->sym;
while (sym && sym->scope == d) {
p->sym = sym->list;
free(sym);
sym = p->sym;
}
return a;
}
expr_t parse(p_t *p, int d) {
char c, *next;
expr_t a, b, sym, expr;
if ((c = *p->input) == '(') {
p->input ++;
next = next_input(p->input);
expr = parse(p, d + 1);
p->input = next;
} else if (IS_LET(p->input)) {
p->input += 4;
expr = parse_let(p, d);
} else if (IS_ADD(p->input)) {
p->input += 4;
a = parse(p, d);
b = parse(p, d);
a = resolve(p, a);
b = resolve(p, b);
expr.type = TYPE_INT;
expr.u.num = a.u.num + b.u.num;
} else if (IS_MULT(p->input)) {
p->input += 5;
a = parse(p, d);
b = parse(p, d);
a = resolve(p, a);
b = resolve(p, b);
expr.type = TYPE_INT;
expr.u.num = a.u.num * b.u.num;
} else if (IS_NUM(p->input) ||
*p->input == '-') {
expr = parse_num(p);
} else {
expr.type = TYPE_ID;
expr.u.id = parse_identifier(p);
}
//assert(*p->input == 0);
return expr;
}
int evaluate(char * expression) {
expr_t result;
p_t p = { 0 };
p.input = expression;
result = parse(&p, 0);
return result.u.num;
}
Input
expression = "(let x 2 (mult x (let x 3 y 4 (add x y))))"
Output
14
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const evaluate = (x) =>
e(
JSON.parse(
x.replace(/[() ]|([a-z][a-z0-9]*)/g, (m) =>
m === '(' ? '[' : m === ')' ? ']' : m === ' ' ? ',' : `"${m}"`
)
)
)
const e = (x, v = []) =>
({
string: () => v.find((y) => y[0] === x)[1],
number: () => x,
object: () =>
({
add: () => e(x[1], v) + e(x[2], v),
mult: () => e(x[1], v) * e(x[2], v),
let: () =>
e(
x[x.length - 1],
x
.slice(1, -1)
.reduce(
({ v, t }, z) =>
t ? { v: [[t, e(z, v)], ...v] } : { v, t: z },
{ v }
).v
),
}[x[0]]()),
}[typeof x]())
Input
expression = "(let x 2 (mult x (let x 3 y 4 (add x y))))"
Output
14
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def evaluate(self, expression):
scopes, items = [{}], [["root"]]
for item in expression.replace(")", " )").split():
if item[0] == "(":
items.append([item[1:]])
if item[1:] == "let":
scopes.append(dict(scopes[-1]))
continue
elif item == ")":
if items[-1][0] == "add":
item = str(int(items[-1][1]) + int(items[-1][-1]))
elif items[-1][0] == "mult":
item = str(int(items[-1][1]) * int(items[-1][-1]))
else:
item = items[-1][-1]
if item in scopes[-1]:
item = scopes[-1][item]
scopes.pop()
items.pop()
if item in scopes[-1] and (items[-1][0] != "let" or len(items[-1]) % 2 == 0):
item = scopes[-1][item]
if items[-1][0] == "let" and item.lstrip("-").isdigit():
scopes[-1][items[-1][-1]] = item
items[-1].append(item)
return int(items[-1][-1])
Input
expression = "(let x 3 x 2 x)"
Output
2