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".
Example 2:
Input: s = "ab##", t = "c#d#" Output: true Explanation: Both s and t become "".
Example 3:
Input: s = "a#c", t = "b" Output: false Explanation: s becomes "c" while t becomes "b".
Constraints:
1 <= s.length, t.length <= 200s and t only contain lowercase letters and '#' characters.Follow up: Can you solve it in O(n) time and O(1) space?
The key idea in #844 Backspace String Compare is to simulate how text editors process the # character, which represents a backspace. After applying all backspaces, we simply check whether the two resulting strings are equal.
A straightforward method uses a stack. Traverse each string and push characters onto the stack. Whenever a # appears, remove the top element if the stack is not empty. After processing both strings, convert the stacks back into strings and compare them. This approach is intuitive and easy to implement.
A more optimized solution uses the two pointers technique. Start from the end of both strings and skip characters that should be deleted due to backspaces. By tracking how many characters must be skipped, we can compare valid characters directly without constructing new strings.
The stack approach uses extra space, while the two-pointer method achieves O(1) space by processing the strings in reverse.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Stack Simulation | O(n + m) | O(n + m) |
| Two Pointers (Reverse Traversal) | O(n + m) | O(1) |
Kevin Naughton Jr.
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
This 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
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The two-pointer technique processes both strings from right to left and skips characters affected by backspaces using counters. Since it does not build intermediate strings or stacks, it only uses a few variables, resulting in constant extra space.
Yes, this problem is a common interview-style question because it tests understanding of string manipulation, stack simulation, and two-pointer optimization. Variations of this problem have appeared in interviews at large tech companies.
A stack is the most intuitive data structure for this problem. It naturally simulates typing behavior by pushing characters and popping them when a backspace appears. While simple, it requires additional memory proportional to the string length.
The optimal approach uses two pointers starting from the end of both strings. By counting pending backspaces and skipping characters that should be deleted, we can directly compare valid characters without building new strings. This reduces space usage to O(1).
In this JavaScript approach, two pointers iterate from the back of each string while adjusting based on backspace characters. By keeping track of effective deletions, we achieve the functional comparison to determine equality of the resultant strings after backspaces are applied.