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 <= 200
• s and t only 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()
    }
}

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