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Formatted question description: https://leetcode.ca/all/2390.html

2390. Removing Stars From a String

  • Difficulty: Medium.
  • Related Topics: String, Stack, Simulation.
  • Similar Questions: Backspace String Compare, Remove All Adjacent Duplicates In String.

Problem

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.

Solution (Java, C++, Python)

  • class Solution {
    public String removeStars(String s) {
        StringBuilder sb = new StringBuilder();
        int stars = 0;
        for (int i = s.length() - 1; i >= 0; --i) {
            if (s.charAt(i) == '*') {
                ++stars;
            } else if (stars > 0) {
                --stars;
            } else {
                sb.append(s.charAt(i));
            }
        }
        return sb.reverse().toString();
    }
}

############

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();
    }
}

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

############

# 2390. Removing Stars From a String
# https://leetcode.com/problems/removing-stars-from-a-string/

class Solution:
    def removeStars(self, s: str) -> str:
        stack = []
        
        for x in s:
            if x == "*":
                stack.pop()
            else:
                stack.append(x)
        
        return "".join(stack)


  • 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;
    }
};

  • 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)
}

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


  • 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
    }
}


  • class Solution {
    /**
     * @param String $s
     * @return String
     */
    function removeStars($s) {
        $rs = [];
        for ($i = 0; $i < strlen($s); $i++) {
            if ($s[$i] == "*") array_pop($rs);
            else array_push($rs, $s[$i]);
        }
        return join($rs);
    }
}

Explain:

nope.

Complexity:

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

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