LeetCode in Kotlin
844. Backspace String Compare
Easy
Given two strings s and t, return true if they are equal when both are typed into empty text editors. '#' means a backspace character.
Note that after backspacing an empty text, the text will continue empty.
Example 1:
Input: s = “ab#c”, t = “ad#c”
Output: true
Explanation: Both s and t become “ac”.
Example 2:
Input: s = “ab##”, t = “c#d#”
Output: true
Explanation: Both s and t become “”.
Example 3:
Input: s = “a#c”, t = “b”
Output: false
Explanation: s becomes “c” while t becomes “b”.
Constraints:
1 <= s.length, t.length <= 200sandtonly contain lowercase letters and'#'characters.
Follow up: Can you solve it in O(n) time and O(1) space?
Solution
class Solution {
fun backspaceCompare(s: String, t: String): Boolean {
return cmprStr(s) == cmprStr(t)
}
private fun cmprStr(str: String): String {
val res = StringBuilder()
var count = 0
for (i in str.length - 1 downTo 0) {
val currChar = str[i]
if (currChar == '#') {
count++
} else {
if (count > 0) {
count--
} else {
res.append(currChar)
}
}
}
return res.toString()
}
}