How do you solve Backspace String Compare on LeetCode?

Backspace String Compare (LeetCode #844) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding how to handle backspaces is crucial.

What is the time complexity of Backspace String Compare?

The optimal solution for Backspace String Compare runs in O(n) time and uses O(1) space. This complexity is linear because we traverse each string once, and we do not use any extra space that grows with input size.

What is the brute force solution for Backspace String Compare?

The brute force approach for Backspace String Compare is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves simulating the typing process for both strings, applying backspaces as we encounter them. This means we will build the final strings character by character and compare them at the end. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Backspace String Compare?

Backspace String Compare uses the following concepts: Two Pointers, String, Stack, Simulation. The recommended patterns to study are: Two Pointers, Stack, String Manipulation.

What are the interview tips for Backspace String Compare?

Always clarify edge cases, like empty strings. Discuss your thought process as you code. Consider space and time complexity early on.

What are common mistakes when solving Backspace String Compare?

Not handling multiple consecutive backspaces correctly. Forgetting to check if the resulting strings are empty.

#844

Backspace String Compare

Easy

Two PointersStringStackSimulationTwo PointersStackString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses two pointers to traverse both strings from the end to the beginning, handling backspaces directly without needing to build new strings. This approach is efficient and avoids unnecessary space usage.

⚙️

Algorithm

3 steps

1Step 1: Initialize two pointers at the end of both strings.
2Step 2: Traverse both strings backwards, skipping characters that are backspaced based on the '#' count.
3Step 3: Compare the characters at both pointers when they are not skipped. If they differ, return false. If both pointers reach the start of the strings simultaneously, return true.

solution.py23 lines

1def backspaceCompare(s: str, t: str) -> bool:
2    def next_valid_index(string, index):
3        skip = 0
4        while index >= 0:
5            if string[index] == '#':
6                skip += 1
7            elif skip > 0:
8                skip -= 1
9            else:
10                break
11            index -= 1
12        return index
13    i, j = len(s) - 1, len(t) - 1
14    while i >= 0 or j >= 0:
15        i = next_valid_index(s, i)
16        j = next_valid_index(t, j)
17        if i >= 0 and j >= 0 and s[i] != t[j]:
18            return False
19        if (i >= 0) != (j >= 0):
20            return False
21        i -= 1
22        j -= 1
23    return True

ℹ

Complexity note: This complexity is linear because we traverse each string once, and we do not use any extra space that grows with input size.

1Understanding how to handle backspaces is crucial.
2Simulating the typing process can help visualize the problem.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.