Backspace-String-Compare

Author: Shikha-code36

README.md

@@ -35,4 +35,5 @@ Check the notes for the explaination - [Notes](https://stingy-shallot-4ea.notion
 
 - [x] [Two Pointers](Two-Pointers)
     - [x] [Squares of a Sorted Array](Two-Pointers/977-Squares-of-a-Sorted-Array.py)
+    - [x] [Backspace String Compare](Two-Pointers/844-Backspace-String-Compare.py)
 

Two-Pointers/844-Backspace-String-Compare.py

@@ -0,0 +1,93 @@
+import itertools
+"Leetcode- https://leetcode.com/problems/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".
+'''
+# Solution-1
+def backspaceCompare(self, S, T):
+    ansS = []
+    for c in S:
+        if c == '#':
+            if ansS:
+                ansS.pop()
+        else:
+            ansS.append(c)
+
+    ansT = []
+    for c in T:
+        if c == '#':
+            if ansT:
+                ansT.pop()
+        else:
+            ansT.append(c)
+
+    return ''.join(ansS) == ''.join(ansT)
+
+    #or#
+
+
+def backspaceCompare(self, S, T):
+    def build(S):
+        ans = []
+        for c in S:
+            if c != '#':
+                ans.append(c)
+            elif ans:
+                ans.pop()
+        return "".join(ans)
+    return build(S) == build(T)
+
+# T:O(M+N)
+# S:O(M+N)
+
+# Solution-2
+def backspaceCompare(self, s, t):
+    l1 = len(s) - 1
+    l2 = len(t) - 1
+
+    while l1 > -1 or l2 > -1:
+        # count how many backspace
+        count = 0
+        while l1 > -1:
+            if s[l1] == "#":
+                count += 1
+                l1 -= 1
+            else:
+                if count == 0:
+                    # not backspace, move on
+                    break
+                else:
+                    # there are backspaces, delete curr char.
+                    l1 -= 1
+                    count -= 1
+
+        count = 0
+        while l2 > -1:
+            if t[l2] == "#":
+                count += 1
+                l2 -= 1
+            else:
+                if count == 0:
+                    break
+                else:
+                    l2 -= 1
+                    count -= 1
+
+        # compare two pointers
+        if l1 > -1 and l2 > -1 and s[l1] != t[l2]:
+            return False
+
+        l1 -= 1
+        l2 -= 1
+    if l1 == l2:
+        return True  
+    else:
+        False