Stack Beats 90%
Intuition\nOur intuition is to simulate the effect of backspaces in the strings and then compare the resulting strings. \nTo do this efficiently, we can use a s
# Intuition\nOur intuition is to simulate the effect of backspaces in the strings and then compare the resulting strings. \nTo do this efficiently, we can use a stack data structure to keep track of the characters after applying the backspaces.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Define a helper function, backspace_run, that takes a string as input and simulates the backspace effect using a stack.\n   - Initialize an empty stack.\n   - Iterate through each character in the input string.\n   - If the character is a regular character, push it onto the stack.\n   - If the character is \'#\', pop the top character from the stack (if the stack is not empty) to simulate the backspace.\n   - Return the resulting stack as a list of characters.\n2. Apply the backspace_run function to both input strings s and t, obtaining final_s and final_t.\n3. Compare final_s and final_t. If they are equal, return True, indicating that the original strings are equal after backspace processing. Otherwise, return False.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity is O(n), where n is the length of the longer of the two input strings. This is because we iterate through the strings once.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity is also O(n) since we use a stack to store characters from the input strings, and the space required is proportional to the length of the input strings.\n\n# Code\n```\nclass Solution:\n    def backspaceCompare(self, s: str, t: str) -> bool:\n        \n        def backspace_run(s):\n            stack = []\n            for char in s:\n                if char == "#":\n                    if stack:\n                        stack.pop()\n                else:\n                    stack.append(char)\n            return stack\n        \n        final_s = backspace_run(s)\n        final_t = backspace_run(t)\n        return final_s == final_t\n```
['Python3']