844_Backspace_String_Compare

Author: qiyuangong

README.md

@@ -151,6 +151,7 @@ Remember solutions are only solutions to given problems. If you want full study
 | 771 | [Jewels and Stones](https://leetcode.com/problems/jewels-and-stones/description/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/771_Jewels_and_Stones.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/771_Jewels_and_Stones.java) | Count given char in string. Hash or table. [Oneline](https://leetcode.com/problems/jewels-and-stones/discuss/113574/1-liners-PythonJavaRuby) |
 | 804 | [Unique Morse Code Words](https://leetcode.com/problems/unique-morse-code-words/description/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/804_Unique_Morse_Code_Words.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/804_Unique_Morse_Code_Words.java) | String, Hash and Set. Set is recommended. |
 | 819 | [Most Common Word](https://leetcode.com/problems/most-common-word/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/819_Most_Common_Word.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/819_Most_Common_Word.java) | String processing, be careful about 'b,b,b'. regex is recommended. |
+| 844 | [Backspace String Compare](https://leetcode.com/problems/backspace-string-compare/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/844_Backspace_String_Compare.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/844_Backspace_String_Compare.java) | 1. Stack pop when encounters #, O(n) and O(n)<br>2. Compare string from end to start, O(n) and O(1) |
 | 904 | [Fruit Into Baskets](https://leetcode.com/problems/fruit-into-baskets/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/904_Fruit_Into_Baskets.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/904_Fruit_Into_Baskets.java) | 1. Scan through blocks of tree, O(n) and O(n)<br>2. Mainten a sliding window with start and curr point, O(n) and O(n).  |
 | 929 | [Unique Email Addresses](https://leetcode.com/problems/unique-email-addresses/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/929_Unique_Email_Addresses.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/929_Unique_Email_Addresses.java) | String handle and hash (or set) |
 | 945 | [Minimum Increment to Make Array Unique](https://leetcode.com/problems/minimum-increment-to-make-array-unique/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/945_Minimum_Increment_to_Make_Array_Unique.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/945_Minimum_Increment_to_Make_Array_Unique.java) | Sort, then list duplicate and missing value in sorted list. O(nlgn) and O(n) |

java/844_Backspace_String_Compare.java

@@ -0,0 +1,29 @@
+class Solution {
+    /*public boolean backspaceCompare(String S, String T) {
+        // https://leetcode.com/problems/backspace-string-compare/discuss/135603/C%2B%2BJavaPython-O(N)-time-and-O(1)-space
+        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;
+        }
+    }*/
+
+    public boolean backspaceCompare(String S, String T) {
+        return trans(S).equals(trans(T));
+    }
+    private String trans(String str) {
+        StringBuilder sb = new StringBuilder();
+        for (char c : str.toCharArray()) {
+            if (c != '#') { sb.append(c); } // if not '#', append it at the end of sb.
+            else if (sb.length() > 0) { sb.deleteCharAt(sb.length() - 1); } // remove last char in sb, if sb is not empty.
+        }
+        return sb.toString();
+    }
+
+}

python/844_Backspace_String_Compare.py

@@ -0,0 +1,49 @@
+class Solution(object):
+    def backspaceCompare(self, S, T):
+        """
+        :type S: str
+        :type T: str
+        :rtype: bool
+        """
+        if S == T:
+            return True
+        s_stack = []
+        t_stack = []
+        for c in S:
+            if c != '#':
+                s_stack.append(c)
+            elif len(s_stack) != 0:
+                s_stack.pop(-1)
+        for c in T:
+            if c != '#':
+                t_stack.append(c)
+            elif len(t_stack) != 0:
+                t_stack.pop(-1)
+        return ''.join(s_stack) == ''.join(t_stack)
+
+    # def backspaceCompare(self, S, T):
+    #     # https://leetcode.com/problems/backspace-string-compare/discuss/135603/C%2B%2BJavaPython-O(N)-time-and-O(1)-space
+    #     back = lambda res, c: res[:-1] if c == '#' else res + c
+    #     return reduce(back, S, "") == reduce(back, T, "")
+
+    # def backspaceCompare(self, S, T):
+    #     def back(res, c):
+    #         if c != '#': res.append(c)
+    #         elif res: res.pop()
+    #         return res
+    #     return reduce(back, S, []) == reduce(back, T, [])
+
+
+    # def backspaceCompare(self, S, T):
+    #     i, j = len(S) - 1, len(T) - 1
+    #     backS = backT = 0
+    #     while True:
+    #         while i >= 0 and (backS or S[i] == '#'):
+    #             backS += 1 if S[i] == '#' else -1
+    #             i -= 1
+    #         while j >= 0 and (backT or T[j] == '#'):
+    #             backT += 1 if T[j] == '#' else -1
+    #             j -= 1
+    #         if not (i >= 0 and j >= 0 and S[i] == T[j]):
+    #             return i == j == -1
+    #         i, j = i - 1, j - 1