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?

class Solution {

    public boolean backspaceCompare(String s, String t) {
        int i = s.length()-1;
        int j = t.length()-1;
        int sSpaceCount = 0;
        int tSpaceCount = 0;

        // While there may be char in String S or String T
        while(i >= 0 || j >= 0) {

            // Find position of next powssible char in String S
            while(i >=0 ) {
                if(s.charAt(i) == '#') {
                    sSpaceCount++;
                    i--;
                } else if(sSpaceCount > 0) {
                    sSpaceCount--;
                    i--;
                }
                else 
                    break;
            }

            // Find position of next powssible char in String T
            while(j >=0 ) {
                if(t.charAt(j) == '#'){
                    tSpaceCount++;
                    j--;
                } else if(tSpaceCount > 0){
                    tSpaceCount--;
                    j--;
                } else 
                    break;
            }

            System.out.println(i+" "+j);

            // If any of the index is negative, then both the strings are not same.
            if(i>=0 != j>=0) 
                return false;

            // If both the index are valid and their characters are not same, then they are not in match
            if( i>=0 && j>=0 && s.charAt(i) != t.charAt(j))
                return false;

            i--;
            j--;
        }

        return true;
    }
}