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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
}
}



Explain:

nope.

Complexity:

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