LeetCode 2390. Removing Stars From a String Solution in Java, C++, Python & More | Explanation + Code

Description

You are given a string s, which contains stars *.

In one operation, you can:

• Choose a star in s.
• Remove the closest non-star character to its left, as well as remove the star itself.

Return the string after all stars have been removed.

Note:

• The input will be generated such that the operation is always possible.
• It can be shown that the resulting string will always be unique.

 

Example 1:

Input: s = "leet**cod*e"
Output: "lecoe"
Explanation: Performing the removals from left to right:
- The closest character to the 1st star is 't' in "leet**cod*e". s becomes "lee*cod*e".
- The closest character to the 2nd star is 'e' in "lee*cod*e". s becomes "lecod*e".
- The closest character to the 3rd star is 'd' in "lecod*e". s becomes "lecoe".
There are no more stars, so we return "lecoe".

Example 2:

Input: s = "erase*****"
Output: ""
Explanation: The entire string is removed, so we return an empty string.

 

Constraints:

• 1 <= s.length <= 105
• s consists of lowercase English letters and stars *.
• The operation above can be performed on s.

Solutions

Solution 1: Stack Simulation

We can use a stack to simulate the operation process. Traverse the string s, and if the current character is not an asterisk, push it onto the stack; if the current character is an asterisk, pop the top element from the stack.

Finally, concatenate the elements in the stack into a string and return it.

The time complexity is O(n), where n is the length of the string s. Ignoring the space consumption of the answer string, the space complexity is O(1).

PythonJavaC++GoTypeScriptRustPHP

class Solution: def removeStars(self, s: str) -> str: ans = [] for c in s: if c == '*': ans.pop() else: ans.append(c) return ''.join(ans)(code-box)

class Solution { public String removeStars(String s) { StringBuilder ans = new StringBuilder(); for (int i = 0; i < s.length(); ++i) { if (s.charAt(i) == '*') { ans.deleteCharAt(ans.length() - 1); } else { ans.append(s.charAt(i)); } } return ans.toString(); } }(code-box)

class Solution { public: string removeStars(string s) { string ans; for (char c : s) { if (c == '*') { ans.pop_back(); } else { ans.push_back(c); } } return ans; } };(code-box)

func removeStars(s string) string { ans := []rune{} for _, c := range s { if c == '*' { ans = ans[:len(ans)-1] } else { ans = append(ans, c) } } return string(ans) }(code-box)

function removeStars(s: string): string { const ans: string[] = []; for (const c of s) { if (c === '*') { ans.pop(); } else { ans.push(c); } } return ans.join(''); }(code-box)

impl Solution { pub fn remove_stars(s: String) -> String { let mut ans = String::new(); for &c in s.as_bytes().iter() { if c == b'*' { ans.pop(); } else { ans.push(char::from(c)); } } ans } }(code-box)

class Solution { /** * @param String $s * @return String */ function removeStars($s) { $ans = []; $n = strlen($s); for ($i = 0; $i < $n; $i++) { $c = $s[$i]; if ($c === '*') { array_pop($ans); } else { $ans[] = $c; } } return implode('', $ans); } }(code-box)