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This approach simulates the typing process using two stacks, one for each string. We iterate over each string, using the stack to build the string as it would appear after considering the backspace characters ('#'). By comparing the final stack representations, we can determine if the two strings are equivalent.
Time Complexity: O(n + m), where n and m are the lengths of s and t. Each character is processed once.
Space Complexity: O(n + m) due to the auxiliary arrays used to store the processed results of s and t.
1def backspaceCompare(s: str, t: str) -> bool:
2 def buildFinalString(string: str) -> str:
3 stack = []
4 for ch in string:
5 if ch != '#':
6 stack.append(ch)
7 elif stack:
8 stack.pop()
9 return ''.join(stack)
10 return buildFinalString(s) == buildFinalString(t)
11
12# Example usage:
13print(backspaceCompare("ab#c", "ad#c")) # True
14print(backspaceCompare("ab##", "c#d#")) # True
15print(backspaceCompare("a#c", "b")) # FalseThis Python solution leverages a list to emulate stack behavior, appending characters unless they're '#', in which case we remove the last element if the list isn't empty. We then compare the two processed strings for equality.
This approach avoids extra space by using two pointers to traverse the strings backwards. By skipping over characters that are effectively backspaced due to a '#' character, we can compare corresponding positions in each string without actually building the resultant strings.
Time Complexity: O(n + m), where n and m are the lengths of s and t.
Space Complexity: O(1), since we only use constant space.
1
The Java solution applies a two-pointer strategy starting from the end of each string. We maintain two skip counters which denote the number of characters to effectively ignore due to backspaces. The pointers skip over these positions and proceed with comparisons only when real characters are encountered.