LeetCode 844 - Backspace String Compare

https://leetcode.com/problems/backspace-string-compare/

Given two strings S and T, return if they are equal when both are typed into empty text editors. # means a backspace character.

Example 1:

Input: S = "ab#c", T = "ad#c"
Output: true
Explanation: Both S and T become "ac".

Approach #2: Two Pointer

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.

See the comments in the code for more details.

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

  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;

  }

https://leetcode.com/problems/backspace-string-compare/discuss/162827/Java-Two-Pointer-With-Explanation-beat-98

    1. Thoughts
        - from back to front
        - if count > 0 && != #, skip
        - if count == 0 && different, return false
        - if #, count++
        - if one ends, the other count == 0 && has any char return false
        - else return true
*/
class Solution {
    public boolean backspaceCompare(String S, String T) {
        int i = S.length()-1;
        int j = T.length()-1;
        int countS = 0;
        int countT = 0;
        while (i >= 0 || j >= 0) {
            while (i >= 0 && (countS > 0 || S.charAt(i) == '#')) {
                if (S.charAt(i) == '#') countS++;
                else countS--;
                i--;
            }
            char left = i < 0 ? '@' : S.charAt(i);
            while (j >= 0 && (countT > 0 || T.charAt(j) == '#')) {
                if (T.charAt(j) == '#') countT++;
                else countT--;
                j--;
            }
            char right = j < 0 ? '@' : T.charAt(j);
            if (left != right) return false;
            i--;
            j--;
        }
        return true;
    }

https://leetcode.com/problems/backspace-string-compare/discuss/135603/C%2B%2BJavaPython-O(N)-time-and-O(1)-space

However, we can do a back string compare (just like the title of problem).
If we do it backward, we meet a char and we can be sure this char won't be deleted.
If we meet a '#', it tell us we need to skip next lowercase char.

    public boolean backspaceCompare(String S, String T) {
        for (int i = S.length() - 1, j = T.length() - 1;; i--, j--) {
            for (int b = 0; i >= 0 && (b > 0 || S.charAt(i) == '#'); --i) b += S.charAt(i) == '#' ? 1 : -1;
            for (int b = 0; j >= 0 && (b > 0 || T.charAt(j) == '#'); --j) b += T.charAt(j) == '#' ? 1 : -1;
            if (i < 0 || j < 0 || S.charAt(i) != T.charAt(j)) return i == -1 && j == -1;
        }
    }

    public boolean backspaceCompare(String S, String T) {
        int i = S.length() - 1, j = T.length() - 1;
        while (true) {
            for (int back = 0; i >= 0 && (back > 0 || S.charAt(i) == '#'); --i)
                back += S.charAt(i) == '#' ? 1 : -1;
            for (int back = 0; j >= 0 && (back > 0 || T.charAt(j) == '#'); --j)
                back += T.charAt(j) == '#' ? 1 : -1;
            if (i >= 0 && j >= 0 && S.charAt(i) == T.charAt(j)) {
                i--; j--;
            } else
                return i == -1 && j == -1;
        }
    }

X.

  public boolean backspaceCompare(String S, String T) {

    return build(S).equals(build(T));

  }

  public String build(String S) {

    Stack<Character> ans = new Stack();

    for (char c : S.toCharArray()) {

      if (c != '#')

        ans.push(c);

      else if (!ans.empty())

        ans.pop();

    }

    return String.valueOf(ans);

  }