30-Day LeetCoding Challenge: Backspace String Compare

The problem

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Solution:

The problem is so easy if you read it carefully it asks if string S and string T are equal consider you typing them in a text editor (notepad), but it is not that easy you will have a special character called # it means backspace so whenever you see it assume you clicked on the backspace button in your keyboard.

So, if you have a string like this ab#c if we follow the rule we mentioned above, you first type a then you find b not a # so type it and it will be ab now we see # and it means backspace so click on the backspace button will remove the b and leave us with a then the last one is c type it ac and this our final result.

It is so obvious now whenever we see the backspace # char we remove the last typed char and this takes to a data structure called Stack as the stack works in LIFO manner which means last-in-first-out.

So, we will go through the two strings to build a stack from their characters and in the end, we compare both of them if they are equal or not.

Let us do an example.

• Example:

S = "ab#c"
T = "ad#c"

We will do each string separately and then combine all elements in the stack into one string and then compare them.

• We will use stack called typed_chars = [].
• Start with S = "ab#c".
• Starting with ch = 'a' is our ch == '#' # backspace no it is not so add it to the stack typed_chars = ['a'].
• With ch = 'b' is ch == '#' no so add it to the stack typed_chars = ['a', 'b'].
• With ch = '#' is ch == '#' yes it is so we simulate the click on the backspace # button by removing the last typed character typed_chars.pop() ==> 'b' so our new stack is typed_chars = ['a'].
• With our last S char ch = 'c' is ch == '#' no it is not so add to the stack typed_chars = ['a', 'c'].

• Build final string from the typed_chars to be new_s = 'ac' remember it.

• Now will do the same with the T = "ad#c" it will be exactly the same process and will end-up with new_t = 'ac'.

• If we compare both new_s == new_t = 'ac' == 'ac' = True, and this is our answer.

• Pseudocode:
  

1. define helper function called helper that takes one string: helper(s)
2.      define typed_chars = []
3.      loop for ch in s:
4.          check if ch != '#':
5.              typed_chars.append(ch)
5.          if ch is equal to # and typed_chars not empty:
6.              typed_chars.pop()
7.      return "".join(typed_chars) to return one string
8. check if helper(S) is equal to helper(T) or not and return the result.

• Complexity:

• Time Complexity:

  • We can see that we loop two times to build our stacks the first one is for O(n) where n is our first-string length, and the second time is for O(m) where m is our second string length, so in total it will be O(n + m).

• Space Complexity:

  • We store our chars in stack two times of course for our first string and our second one, so in total it will be O(n + m).

• Solution in python