LeetCode 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 = "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

Stack based solution

class Solution {
    public boolean backspaceCompare(String s, String t) {
        // create two stacks
        Stack<Character> stk1 = new Stack<>();
        Stack<Character> stk2 = new Stack<>();
        
        // add each element to the stack and pop elements if the stack is not empty and value is'#'
        // otherwise add elements to the stack
        for(int i = 0; i < s.length(); i++){
            if(s.charAt(i) == '#') {
                if(!stk1.isEmpty()) stk1.pop();
            } else {
                stk1.push(s.charAt(i));
            }
        }
        
        for(int i=0; i< t.length(); i++){
             if(t.charAt(i) == '#') {
                if(!stk2.isEmpty()) stk2.pop();
            } else {
                stk2.push(t.charAt(i));
            }  
        }
        
        // is stack size doesnt match, return false
        if(stk1.size() != stk2.size()) return false;
        
        // pop each element from the stack and compare the values
        while(!stk1.isEmpty() && !stk2.isEmpty()){
            if(stk1.pop() != stk2.pop()) return false;
        }
        return true;
    }
}

• Time Complexity: O(M + N)O(M+N), where M, NM,N are the lengths of S and T respectively.

• Space Complexity: O(M + N)O(M+N).

Two Pointer

Intuition

When writing a character, it may or may not be part of the final string depending on how many backspace keystrokes occur in the future.

If instead, we iterate through the string in reverse, then we will know how many backspace characters we have seen, and therefore whether the result includes our character.

Algorithm

Iterate through the string in reverse. If we see a backspace character, the next non-backspace character is skipped. If a character isn't skipped, it is part of the final answer.

class Solution {
    public boolean backspaceCompare(String S, String T) {
        int i = S.length() - 1, j = T.length() - 1;
        int skipS = 0, skipT = 0;

        while (i >= 0 || j >= 0) { // While there may be chars in build(S) or build (T)
            while (i >= 0) { // Find position of next possible char in build(S)
                if (S.charAt(i) == '#') {skipS++; i--;}
                else if (skipS > 0) {skipS--; i--;}
                else break;
            }
            while (j >= 0) { // Find position of next possible char in build(T)
                if (T.charAt(j) == '#') {skipT++; j--;}
                else if (skipT > 0) {skipT--; j--;}
                else break;
            }
            // If two actual characters are different
            if (i >= 0 && j >= 0 && S.charAt(i) != T.charAt(j))
                return false;
            // If expecting to compare char vs nothing
            if ((i >= 0) != (j >= 0))
                return false;
            i--; j--;
        }
        return true;
    }
}