LeetCode-in-Java

3612. Process String with Special Operations I

Medium

You are given a string s consisting of lowercase English letters and the special characters: *, #, and %.

Build a new string result by processing s according to the following rules from left to right:

• If the letter is a lowercase English letter append it to result.
• A '*' removes the last character from result, if it exists.
• A '#' duplicates the current result and appends it to itself.
• A '%' reverses the current result.

Return the final string result after processing all characters in s.

Example 1:

Input: s = “a#b%*”

Output: “ba”

Explanation:

i	s[i]	Operation	Current result
0	'a'	Append 'a'	"a"
1	'#'	Duplicate result	"aa"
2	'b'	Append 'b'	"aab"
3	'%'	Reverse result	"baa"
4	'*'	Remove the last character	"ba"

Thus, the final result is "ba".

Example 2:

Input: s = “z*#”

Output: “”

Explanation:

i	s[i]	Operation	Current result
0	'z'	Append 'z'	"z"
1	'*'	Remove the last character	""
2	'#'	Duplicate the string	""

Thus, the final result is "".

Constraints:

• 1 <= s.length <= 20
• s consists of only lowercase English letters and special characters *, #, and %.

Solution

public class Solution {
    public String processStr(String s) {
        StringBuilder res = new StringBuilder();
        for (char c : s.toCharArray()) {
            if (c != '*' && c != '#' && c != '%') {
                res.append(c);
            } else if (c == '#') {
                res.append(res);
            } else if (c == '%') {
                res.reverse();
            } else {
                if (!res.isEmpty()) {
                    res.deleteCharAt(res.length() - 1);
                }
            }
        }
        return res.toString();
    }
}

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