LeetCode 844 - Backspace String Compare

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

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".

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".

Note:

1. 1 <= S.length <= 200
2. 1 <= T.length <= 200
3. S and T only contain lowercase letters and '#' characters.

Follow up:

• Can you solve it in O(N) time and O(1) space?

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.

https://leetcode.com/problems/backspace-string-compare/discuss/135603/C++JavaPython-4-lines-O(N)-time-and-O(1)-space

    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;
        }
    }

https://blog.csdn.net/qjh5606/article/details/81509282

bool backspaceCompare(string S, string T) {

  int i = S.size() - 1, j = T.size() - 1, countA = 0, countB = 0;

  while (i >= 0 || j >= 0) {

      while (i >= 0 && (S[i] == '#' || countA > 0)) 

          S[i--] == '#' ? countA++ : countA--;

      while (j >= 0 && (T[j] == '#' || countB > 0)) 

          T[j--] == '#' ? countB++ : countB--;

      if (i < 0 || j < 0) 

          return i == j;

      if (S[i--] != T[j--]) 

          return false;

  }

  return i == j;

}

https://whutweihuan.github.io/2018/09/08/2018-9-8-blog_leetcode3/

    bool 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 (i >= 0) {
                if (S[i] == '#') { skipS++; i--; }
                else if (skipS > 0) { skipS--; i--; }
                else { break; }
            }
            while (j >= 0) {
                if (T[j] == '#') { skipT++;j--; }
                else if (skipT > 0) { skipT--; j--; }
                else { break; }
            }
            if (i>=0 && j>=0 && S[i] != T[j]) { return false; }
            i--, j--;
        }
        return false;
    }

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

  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;

  }

  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);

  }

X.  https://zhanghuimeng.github.io/post/leetcode-844-backspace-string-compare/

    bool backspaceCompare(string S, string T) {
        int topS = 0, topT = 0;
        for (int i = 0; i < S.length(); i++) {
            char c = S[i];
            if (c == '#') {
                if (topS > 0)
                    topS--;
            }
            else
                S[topS++] = c;
        }
        for (int i = 0; i < T.length(); i++) {
            char c = T[i];
            if (c == '#') {
                if (topT > 0)
                    topT--;
            }
            else
                T[topT++] = c;
        }

        if (topS != topT)
            return false;
        return S.substr(0, topS) == T.substr(0, topT);
    }