844. Backspace String Compare

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 = "#a#c"
Output: true
Explanation: Both s and t become "c".

Example 4:

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

Compare from the rear.

Keep delete Count for each string

public boolean backspaceCompare(String s, String t) {
    int deleteS = 0;
    int deleteT = 0;
    int i = s.length()-1, j = t.length()-1;
    while(true){

       while(i >= 0 && (deleteS > 0 || s.charAt(i) == '#')){
           deleteS += s.charAt(i) == '#' ? 1 : -1;
           i--;
       }
       while(j >= 0 && (deleteT > 0 || t.charAt(j) == '#')){
           deleteT += t.charAt(j) == '#' ? 1 : -1;
           j--;
       }
       if(i>=0 && j >= 0 && s.charAt(i) == t.charAt(j)){
           i--;
           j--;
       }else {
            break;
       }
   }
   return i == -1 && j == -1;
}