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3614. Process String with Special Operations II
Description
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 fromresult, if it exists. - A
'#'duplicates the currentresultand appends it to itself. - A
'%'reverses the currentresult.
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 <= 105sconsists of only lowercase English letters and special characters'*','#', and'%'.0 <= k <= 1015- The length of
resultafter processingswill not exceed1015.