parser Archives - Huahua's Tech Road

花花酱 LeetCode 1678. Goal Parser Interpretation

By zxi on December 5, 2020

You own a Goal Parser that can interpret a string command. The command consists of an alphabet of "G", "()" and/or "(al)" in some order. The Goal Parser will interpret "G" as the string "G", "()" as the string "o", and "(al)" as the string "al". The interpreted strings are then concatenated in the original order.

Given the string command, return the Goal Parser‘s interpretation of command.

Example 1:

Input: command = "G()(al)"
Output: "Goal"
Explanation: The Goal Parser interprets the command as follows:
G -> G
() -> o
(al) -> al
The final concatenated result is "Goal".

Example 2:

Input: command = "G()()()()(al)"
Output: "Gooooal"

Example 3:

Input: command = "(al)G(al)()()G"
Output: "alGalooG"

Constraints:

1 <= command.length <= 100
command consists of "G", "()", and/or "(al)" in some order.

Solution: String

If we encounter ‘(‘ check the next character to determine whether it’s ‘()’ or ‘(al’)

Time complexity: O(n)
Space complexity: O(n)

C++

class Solution {

public:

string interpret(string command) {

string ans;

for (int i = 0; i < command.size(); ++i) {

if (command[i] == 'G') ans += "G";

else if (command[i] == '(') {

if (command[i + 1] == ')') ans += "o";

else ans += "al";

}

return ans;

}

};

Python3

class Solution:

def interpret(self, command: str) -> str:

ans = []

for i, c in enumerate(command):

if c == 'G': ans.append('G')

elif c == '(':

ans.append('o' if command[i + 1] == ')' else 'al')

return ''.join(ans)

花花酱 LeetCode 1106. Parsing A Boolean Expression

By zxi on June 30, 2019

Return the result of evaluating a given boolean expression, represented as a string.

An expression can either be:

"t", evaluating to True;
"f", evaluating to False;
"!(expr)", evaluating to the logical NOT of the inner expression expr;
"&(expr1,expr2,...)", evaluating to the logical AND of 2 or more inner expressions expr1, expr2, ...;
"|(expr1,expr2,...)", evaluating to the logical OR of 2 or more inner expressions expr1, expr2, ...

Example 1:

Input: expression = "!(f)"
Output: true

Example 2:

Input: expression = "|(f,t)"
Output: true

Example 3:

Input: expression = "&(t,f)"
Output: false

Example 4:

Input: expression = "|(&(t,f,t),!(t))"
Output: false

Solution: Recursion

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

bool parseBoolExpr(string expression) {

int s = 0;

return parse(expression, s);

}

private:

bool parse(const string& exp, int& s) {

char ch = exp[s++];

if (ch == 't') return true;

if (ch == 'f') return false;

if (ch == '!') {

bool ans = !parse(exp, ++s);

++s;

return ans;

}

bool is_and = (ch == '&');

bool ans = is_and;

++s;

while (true) {

if (is_and) ans &= parse(exp, s);

else ans |= parse(exp, s);

if (exp[s++] == ')') break;

}

return ans;

}

};

Java

class Solution {

private int s;

private char[] exp;

public boolean parseBoolExpr(String expression) {

exp = expression.toCharArray();

s = 0;

return parse() == 1;

}

private int parse() {

char ch = exp[s++];

if (ch == 't') return 1;

if (ch == 'f') return 0;

if (ch == '!') {

++s;

int ans = 1 - parse();

++s;

return ans;

}

boolean is_and = (ch == '&');

int ans = is_and ? 1 : 0;

++s;

while (true) {

if (is_and) ans &= parse();

else ans |= parse();

if (exp[s++] == ')') break;

}

return ans;

}

花花酱 LeetCode 736. Parse Lisp Expression

By zxi on December 1, 2017

Problem:

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, a let-expression, an add-expression, a 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 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 v2 is 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 expr.
An add-expression takes the form (add e1 e2) where add is always the string "add", there are always two expressions e1, e2, and this expression evaluates to the addition of the evaluation of e1 and the evaluation of e2.
A mult-expression takes the form (mult e1 e2) where mult is always the string "mult", there are always two expressions e1, e2, and this expression evaluates to the multiplication of the evaluation of e1 and the evaluation of e2.
For the purposes of 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”, or “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 scope.

Evaluation Examples:

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.

Note:

The given string expression is 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.
The length of expression is at most 2000. (It is also non-empty, as that would not be a legal expression.)
The answer and all intermediate calculations of that answer are guaranteed to fit in a 32-bit integer.

Idea:

Recursive parsing

Time complexity: O(n^2) in worst case O(n) in practice

Space complexity: O(n)

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

int evaluate(string expression) {

scopes_.clear();

int pos = 0;

return eval(expression, pos);

}

private:

int eval(const string& s, int& pos) {

scopes_.push_front(unordered_map<string, int>());

int value = 0; // The return value of current expr

if (s[pos] == '(') ++pos;

// command, variable or number

const string token = getToken(s, pos);

if (token == "add") {

int v1 = eval(s, ++pos);

int v2 = eval(s, ++pos);

value = v1 + v2;

} else if (token == "mult") {

int v1 = eval(s, ++pos);

int v2 = eval(s, ++pos);

value = v1 * v2;

} else if (token == "let") {

string var;

// expecting " var1 exp1 var2 exp2 ... last_expr)"

while (s[pos] != ')') {

++pos;

// Must be last_expr

if (s[pos] == '(') {

value = eval(s, ++pos);

break;

}

// Get a token, could be "x" or "-12" for last_expr

var = getToken(s, pos);

// End of let, var is last_expr

if (s[pos] == ')') {

if (isalpha(var[0]))

value = getValue(var);

else

value = stoi(var);

break;

}

// x -12 -> set x to -12 and store in the current scope and take it as the current return value

value = scopes_.front()[var] = eval(s, ++pos);

}

} else if (isalpha(token[0])) {

value = getValue(token); // symbol

} else {

value = std::stoi(token); // number

}

if (s[pos] == ')') ++pos;

scopes_.pop_front();

return value;

}

int getValue(const string& symbol) {

for (const auto& scope : scopes_)

if (scope.count(symbol)) return scope.at(symbol);

return 0;

}

// Get a token from current pos.

// "let x" -> "let"

// "-12 (add x y)" -> "-12"

string getToken(const string& s, int& pos) {

string token;

while (pos < s.length()) {

if (s[pos] == ')' || s[pos] == ' ') break;

token += s[pos++];

}

return token;

}

deque<unordered_map<string, int>> scopes_;

};

Posts tagged as “parser”

花花酱 LeetCode 1678. Goal Parser Interpretation

Solution: String

C++

Python3

花花酱 LeetCode 1106. Parsing A Boolean Expression

Solution: Recursion

C++

Java

花花酱 LeetCode 736. Parse Lisp Expression