LeetCode-in-Java

3614. Process String with Special Operations II

Hard

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

You are also given an integer k.

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 kth character of the final string result. If k is out of the bounds of result, return '.'.

Example 1:

Input: s = “a#b%*”, k = 1

Output: “a”

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"

The final result is "ba". The character at index k = 1 is 'a'.

Example 2:

Input: s = “cd%#*#”, k = 3

Output: “d”

Explanation:

i	s[i]	Operation	Current result
0	'c'	Append 'c'	"c"
1	'd'	Append 'd'	"cd"
2	'%'	Reverse result	"dc"
3	'#'	Duplicate result	"dcdc"
4	'*'	Remove the last character	"dcd"
5	'#'	Duplicate result	"dcddcd"

The final result is "dcddcd". The character at index k = 3 is 'd'.

Example 3:

Input: s = “z*#”, k = 0

Output: “.”

Explanation:

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

The final result is "". Since index k = 0 is out of bounds, the output is '.'.

Constraints:

• 1 <= s.length <= 105
• s consists of only lowercase English letters and special characters '*', '#', and '%'.
• 0 <= k <= 1015
• The length of result after processing s will not exceed 1015.

Solution

public class Solution {
    public char processStr(String s, long k) {
        long len = 0;
        for (char c : s.toCharArray()) {
            if (Character.isLowerCase(c)) {
                len++;
            } else if (c == '*' && len > 0) {
                len--;
            } else if (c == '#') {
                len *= 2;
            }
        }
        if (k >= len) {
            return '.';
        }
        for (int i = s.length() - 1; i >= 0; i--) {
            char c = s.charAt(i);
            if (Character.isLowerCase(c)) {
                if (k == len - 1) {
                    return c;
                }
                len--;
            } else if (c == '*') {
                len++;
            } else if (c == '#') {
                len /= 2;
                if (k >= len) {
                    k -= len;
                }
            } else if (c == '%') {
                k = len - 1 - k;
            }
        }
        return '.';
    }
}

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