What is the time complexity of Backspace String Compare?

The optimal solution has a time complexity of O(m+n).

What is the space complexity of Backspace String Compare?

The space complexity is O(m+n).

What data structures are used to solve Backspace String Compare?

This problem uses Two Pointers, String, Stack, Simulation. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Backspace String Compare?

Insight 1: The core idea is to simulate the backspace operations and compare the resulting strings. Backspaces cancel out characters to their left, so we need to process the string in reverse order to easily handle backspaces. Insight 2: A two-pointer approach is efficient for comparing the strings while simultaneously handling backspaces without explicitly constructing the processed strings in memory, allowing O(1) space complexity.

Which companies ask LeetCode 844. Backspace String Compare in interviews?

This problem is frequently asked at Agoda, Amdocs, Booking.com, Chewy, Grammarly and 5 more companies.

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Backspace String Compare - LeetCode 844 Solution

Backspace String Compare - Complete Solution Guide

Backspace String Compare is LeetCode problem 844, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp.

Problem Statement

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". Constra

Detailed Explanation

The problem asks us to compare two strings, `s` and `t`, and determine if they are equal after applying backspace operations. A backspace character, represented by '#', deletes the character immediately to its left (if any). We need to return `true` if the processed strings are equal and `false` otherwise. Importantly, backspacing an empty string has no effect.

Solution Approach

The Java and C++ solutions employ a two-pointer approach, iterating from the end of each string. They maintain a `backspace` counter for each string, incrementing it when '#' is encountered and decrementing it when a non-'#' character is encountered while `backspace` > 0. The characters are compared only when the `backspace` counter is zero for both strings. The Python solution uses a stack to build the processed string.

Step-by-Step Algorithm

Step 1: Initialize two pointers, `i` and `j`, to the last index of strings `s` and `t`, respectively.
Step 2: Initialize two `backspace` counters, `backspaceS` and `backspaceT`, to 0.
Step 3: While either `i` or `j` is within the bounds of their respective strings, iterate and process each character.
Step 4: For each string (`s` and `t`), move the pointer backward until either a non-'#' character with `backspace` count of 0 is encountered, or the start of the string is reached.
Step 5: If the current character is '#', increment the respective `backspace` counter and move the pointer back.
Step 6: If the current character is not '#' and the `backspace` counter is greater than 0, decrement the `backspace` counter and move the pointer back.
Step 7: Once both pointers point to valid characters (or reach the beginning of the strings) after skipping backspaces, compare the characters at `s[i]` and `t[j]`.
Step 8: If the characters are different, return `false`.
Step 9: If one string has remaining characters after skipping backspaces while the other doesn't, return `false`.
Step 10: Move both pointers one step backward.
Step 11: If the loop completes without finding any differences, return `true`.

Key Insights

Insight 1: The core idea is to simulate the backspace operations and compare the resulting strings. Backspaces cancel out characters to their left, so we need to process the string in reverse order to easily handle backspaces.
Insight 2: A two-pointer approach is efficient for comparing the strings while simultaneously handling backspaces without explicitly constructing the processed strings in memory, allowing O(1) space complexity.
Insight 3: Multiple consecutive '#' characters need to be accounted for, as each cancels out a character to the left.

Complexity Analysis

Time Complexity: O(m+n)

Space Complexity: O(m+n)

Topics

This problem involves: Two Pointers, String, Stack, Simulation.

Companies

Asked at: Agoda, Amdocs, Booking.com, Chewy, Grammarly, IBM, Microstrategy, Roku, Tinkoff, Wayfair.

Backspace String Compare | LeetCode 844

Easy

Two Pointers String Stack Simulation

Time: O(m+n)Space: O(m+n)

Solution Code

Available in:

class Solution {
    public boolean backspaceCompare(String s, String t) {
        int i = s.length() - 1;
        int j = t.length() - 1;
        int backspaceS = 0;
        int backspaceT = 0;

        while (i >= 0 || j >= 0) {
            while (i >= 0) {
                if (s.charAt(i) == '#') {
                    backspaceS++;
                    i--;
                } else if (backspaceS > 0) {
                    backspaceS--;
                    i--;
                } else {
                    break;
                }
            }
            while (j >= 0) {
                if (t.charAt(j) == '#') {
                    backspaceT++;
                    j--;
                } else if (backspaceT > 0) {
                    backspaceT--;
                    j--;
                } else {
                    break;
                }
            }

            if (i >= 0 && j >= 0 && s.charAt(i) == t.charAt(j)) {
                i--;
                j--;
            } else if (i >= 0 || j >= 0) {
                return false;
            }
        }
        return true;
    }
}

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Problem Info

Description

Problem 844: Backspace String Compare - Detailed explanation and solution approaches.

Topics

Two Pointers String Stack Simulation

Companies

AgodaAmdocsBooking.comChewyGrammarlyIBMMicrostrategyRoku

Stats

Difficulty:Easy

Languages:3

Companies:10

Problem #:844

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Initialize two pointers, `i` and `j`, to the last index of strings `s` and `t`, respectively.
2
Step 2: Initialize two `backspace` counters, `backspaceS` and `backspaceT`, to 0.
3
Step 3: While either `i` or `j` is within the bounds of their respective strings, iterate and process each character.
4
Step 4: For each string (`s` and `t`), move the pointer backward until either a non-'#' character with `backspace` count of 0 is encountered, or the start of the string is reached.
5
Step 5: If the current character is '#', increment the respective `backspace` counter and move the pointer back.
6
Step 6: If the current character is not '#' and the `backspace` counter is greater than 0, decrement the `backspace` counter and move the pointer back.
7
Step 7: Once both pointers point to valid characters (or reach the beginning of the strings) after skipping backspaces, compare the characters at `s[i]` and `t[j]`.
8
Step 8: If the characters are different, return `false`.
9
Step 9: If one string has remaining characters after skipping backspaces while the other doesn't, return `false`.
10
Step 10: Move both pointers one step backward.
11
Step 11: If the loop completes without finding any differences, return `true`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m+n)

💾

Space Complexity

O(m+n)

📝

Why?

The two-pointer approach in Java and C++ modifies the pointers in place without creating extra space. The Python solution explicitly builds the strings which results in O(m+n) space complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using Stack

Iterate through each string, pushing characters onto a stack and popping characters from the stack when a '#' is encountered. Construct the final string from the remaining elements in the stack.

Time: O(m+n)Space: O(m+n)

Trade-offs: This approach is easier to understand than the two-pointer approach, but it uses O(m+n) space. The two-pointer approach is preferred when space is a concern.

✨

Key Insights

1Insight 1: The core idea is to simulate the backspace operations and compare the resulting strings. Backspaces cancel out characters to their left, so we need to process the string in reverse order to easily handle backspaces.
2Insight 2: A two-pointer approach is efficient for comparing the strings while simultaneously handling backspaces without explicitly constructing the processed strings in memory, allowing O(1) space complexity.
3Insight 3: Multiple consecutive '#' characters need to be accounted for, as each cancels out a character to the left.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty strings - The algorithm correctly handles empty strings because the while loop condition `i >= 0 || j >= 0` handles the case where both strings are empty or one is empty and other has backspaces. The `backspace` counters prevent reading out of bounds.
Edge case 2: String with only '#' characters - The algorithm correctly handles strings with only '#' characters because it will iterate through all the '#' characters and decrement the pointer appropriately. The loops will correctly terminate once the beginning of the string is reached.
Edge case 3: Consecutive '#' characters - The algorithm handles consecutive '#' characters by incrementing the backspace counter appropriately. So all characters to the left will be skipped.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling consecutive '#' characters by only skipping one character instead of the correct number based on backspace count.

✗Common mistake 2: Not considering the case when one string has been fully processed but the other still has characters remaining (after backspace processing).

✗Common mistake 3: Using an auxiliary string with additional space instead of the optimal O(1) two-pointer solution.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Text editors with backspace functionality. This algorithm can be used to simulate and compare the final text content after backspace operations.

🏦Real-world scenario 2: Command-line interfaces. Command-line interpreters often use backspace functionality. This logic can be applied to correctly process and compare commands entered by a user.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

The two-pointer technique is powerful for sorted arrays. The key insight is recognizing when elements can be processed from both ends simultaneously.

🎯 Interview Tip

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

The most common mistake is not handling the case when pointers meet. Always consider the termination condition.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 3, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Backspace String Compare"?

For this easy-level Two Pointers problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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