Move from brute-force thinking to an efficient approach using stack strategy.
You are given a string s, which contains stars *.
In one operation, you can:
s.Return the string after all stars have been removed.
Note:
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 <= 105s consists of lowercase English letters and stars *.s.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Stack
"leet**cod*e"
"erase*****"
backspace-string-compare)remove-all-adjacent-duplicates-in-string)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2390: Removing Stars From a String
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();
}
}
// Accepted solution for LeetCode #2390: Removing Stars From a String
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)
}
# Accepted solution for LeetCode #2390: Removing Stars From a String
class Solution:
def removeStars(self, s: str) -> str:
ans = []
for c in s:
if c == '*':
ans.pop()
else:
ans.append(c)
return ''.join(ans)
// Accepted solution for LeetCode #2390: Removing Stars From a String
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
}
}
// Accepted solution for LeetCode #2390: Removing Stars From a String
function removeStars(s: string): string {
const ans: string[] = [];
for (const c of s) {
if (c === '*') {
ans.pop();
} else {
ans.push(c);
}
}
return ans.join('');
}
Use this to step through a reusable interview workflow for this problem.
For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.
Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.
Review these before coding to avoid predictable interview regressions.
Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.
Usually fails on: Indices point to blocked elements and outputs shift.
Fix: Pop while invariant is violated before pushing current element.