Removing Stars From a String

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.

Code

Stack

Time Complexity: $O(N)$, Space Complexity: $O(N)$

class Solution {
public:
    string removeStars(string s) {
        string answer = "";
        stack<char> st;
        for(int i = 0; i < s.length(); i++) {
            if(s[i] != '*') st.push(s[i]);
            else st.pop();
        }
 
        while(!st.empty()) {
            answer += st.top(); st.pop();
        }
 
        reverse(answer.begin(), answer.end());
        return answer;
        
    }
};

Two pointer

用 two pointer 模擬 stack 的 pop & push 相對位置（index）。

Time Complexity: $O(N)$, Space Complexity: $O(1)$

class Solution {
public:
    string removeStars(string s) {
        
        int j = 0;
        for(int i = 0; i < s.length(); i++) {
            if(s[i] != '*') {
                s[j++] = s[i];
            } else {
                j--;
            };
        }
 
        return s.substr(0, j);
        
    }
};

Source

• Removing Stars From a String - LeetCode