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:
result.'*' removes the last character from result, if it exists.'#' duplicates the current result and appends it to itself.'%' 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 <= 105s consists of only lowercase English letters and special characters '*', '#', and '%'.0 <= k <= 1015result after processing s will not exceed 1015.class Solution {
public char processStr(String s, long k) {
int n = s.length();
long len = 0;
for (int i = 0; i < n; i++) {
char c = s.charAt(i);
switch (c) {
case '*' -> {
if (len > 0)
len--;
}
case '#' -> len *= 2;
case '%' -> {
}
default -> len++;
}
}
if (k + 1 > len)
return '.';
for (int i = n - 1; i >= 0; i--) {
char c = s.charAt(i);
switch (c) {
case '*' -> len++;
case '#' -> {
if (k + 1 > (len + 1) / 2)
k -= len / 2;
len = (len + 1) / 2;
}
case '%' -> k = len - k - 1;
default -> {
if (k + 1 == len)
return c;
len--;
}
}
}
return '.';
}
}