Description
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 <= 100command consists of "G", "()", and/or "(al)" in some order.
Solutions
Solution 1: String Replacement
According to the problem, we only need to replace "()" with 'o' and "(al)" with "al" in the string command.
PythonJavaC++GoTypeScriptRustC
class Solution:
def interpret(self, command: str) -> str:
return command.replace('()', 'o').replace('(al)', 'al')(code-box)
class Solution {
public String interpret(String command) {
return command.replace("()", "o").replace("(al)", "al");
}
}(code-box)
class Solution {
public:
string interpret(string command) {
while (command.find("()") != -1) command.replace(command.find("()"), 2, "o");
while (command.find("(al)") != -1) command.replace(command.find("(al)"), 4, "al");
return command;
}
};(code-box)
func interpret(command string) string {
command = strings.ReplaceAll(command, "()", "o")
command = strings.ReplaceAll(command, "(al)", "al")
return command
}(code-box)
function interpret(command: string): string {
return command.replace(/\(\)/g, 'o').replace(/\(al\)/g, 'al');
}(code-box)
impl Solution {
pub fn interpret(command: String) -> String {
command.replace("()", "o").replace("(al)", "al")
}
}(code-box)
char* interpret(char* command) {
int n = strlen(command);
char* ans = malloc(sizeof(char) * n + 1);
int i = 0;
for (int j = 0; j < n; j++) {
char c = command[j];
if (c == 'G') {
ans[i++] = 'G';
} else if (c == '(') {
if (command[j + 1] == ')') {
ans[i++] = 'o';
} else {
ans[i++] = 'a';
ans[i++] = 'l';
}
}
}
ans[i] = '\0';
return ans;
}(code-box)
Solution 2: String Iteration
We can also iterate over the string command. For each character c:
- If it is
'G', directly add c to the result string;
- If it is
'(', check if the next character is ')'. If it is, add 'o' to the result string. Otherwise, add "al" to the result string.
After the iteration, return the result string.
The time complexity is O(n), and the space complexity is O(1).
PythonJavaC++GoTypeScriptRust
class Solution:
def interpret(self, command: str) -> str:
ans = []
for i, c in enumerate(command):
if c == 'G':
ans.append(c)
elif c == '(':
ans.append('o' if command[i + 1] == ')' else 'al')
return ''.join(ans)(code-box)
class Solution {
public String interpret(String command) {
StringBuilder ans = new StringBuilder();
for (int i = 0; i < command.length(); ++i) {
char c = command.charAt(i);
if (c == 'G') {
ans.append(c);
} else if (c == '(') {
ans.append(command.charAt(i + 1) == ')' ? "o" : "al");
}
}
return ans.toString();
}
}(code-box)
class Solution {
public:
string interpret(string command) {
string ans;
for (int i = 0; i < command.size(); ++i) {
char c = command[i];
if (c == 'G')
ans += c;
else if (c == '(')
ans += command[i + 1] == ')' ? "o" : "al";
}
return ans;
}
};(code-box)
func interpret(command string) string {
ans := &strings.Builder{}
for i, c := range command {
if c == 'G' {
ans.WriteRune(c)
} else if c == '(' {
if command[i+1] == ')' {
ans.WriteByte('o')
} else {
ans.WriteString("al")
}
}
}
return ans.String()
}(code-box)
function interpret(command: string): string {
const n = command.length;
const ans: string[] = [];
for (let i = 0; i < n; i++) {
const c = command[i];
if (c === 'G') {
ans.push(c);
} else if (c === '(') {
ans.push(command[i + 1] === ')' ? 'o' : 'al');
}
}
return ans.join('');
}(code-box)
impl Solution {
pub fn interpret(command: String) -> String {
let mut ans = String::new();
let bs = command.as_bytes();
for i in 0..bs.len() {
if bs[i] == b'G' {
ans.push_str("G");
}
if bs[i] == b'(' {
ans.push_str({
if bs[i + 1] == b')' {
"o"
} else {
"al"
}
});
}
}
ans
}
}(code-box)
