What data structures are used to solve Apply Bitwise Operations to Make Strings Equal?

This problem uses String, Bit Manipulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Apply Bitwise Operations to Make Strings Equal?

Insight 1: The crucial insight is recognizing that the presence or absence of '1's in the string is invariant under the given operation. The total number of '1's may change, but if there's at least one '1' initially, the operation can propagate it. If there are no '1's initially, the operation cannot create one. Insight 2: The `OR` and `XOR` operations, when applied together in the specified manner, ensure that the total number of bits set to '1' are either preserved or potentially increased. The presence of even one '1' in the string allows us to potentially set any other bit to '1' via the described operation.

Which companies ask LeetCode 2546. Apply Bitwise Operations to Make Strings Equal in interviews?

This problem is frequently asked at Sprinklr.

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Apply Bitwise Operations to Make Strings Equal - LeetCode 2546 Solution

Q: How to solve Apply Bitwise Operations to Make Strings Equal?

The provided solution directly leverages the key insight. It checks if the presence of at least one '1' is the same in both strings. This is done by simply checking if '1' exists within each string using the `in` operator. If both strings either have at least one '1' or neither has any '1's, it means we can transform `s` to `target`, and the function returns `True`. Otherwise, it returns `False`.

Q: What is the time complexity of Apply Bitwise Operations to Make Strings Equal?

The optimal solution has a time complexity of O(n), where n is the length of the string. The `in` operator in Python has a time complexity of O(n) in the worst case, as it might need to scan the entire string..

Q: What is the space complexity of Apply Bitwise Operations to Make Strings Equal?

The space complexity is O(1). The solution uses a constant amount of extra space, regardless of the input size..

Apply Bitwise Operations to Make Strings Equal - Complete Solution Guide

Apply Bitwise Operations to Make Strings Equal is LeetCode problem 2546, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3.

Problem Statement

You are given two 0-indexed binary strings s and target of the same length n . You can do the following operation on s any number of times: Choose two different indices i and j where 0 <= i, j < n . Simultaneously, replace s[i] with ( s[i] OR s[j] ) and s[j] with ( s[i] XOR s[j] ). For example, if s = "0110" , you can choose i = 0 and j = 2 , then simultaneously replace s[0] with ( s[0] OR s[2] = 0 OR 1 = 1 ), and s[2] with ( s[0] XOR s[2] = 0 XOR 1 = 1 ), so we will have s = "1110" . Return tru

Detailed Explanation

The problem asks whether we can transform a binary string `s` into another binary string `target` of the same length by repeatedly applying a specific bitwise operation. The operation involves selecting two distinct indices `i` and `j`, and simultaneously updating `s[i]` with `s[i] OR s[j]` and `s[j]` with `s[i] XOR s[j]`. The goal is to determine if any number of these operations can make `s` identical to `target`.

Solution Approach

The provided solution directly leverages the key insight. It checks if the presence of at least one '1' is the same in both strings. This is done by simply checking if '1' exists within each string using the `in` operator. If both strings either have at least one '1' or neither has any '1's, it means we can transform `s` to `target`, and the function returns `True`. Otherwise, it returns `False`.

Step-by-Step Algorithm

Step 1: Check if the character '1' is present in string `s`.
Step 2: Check if the character '1' is present in string `target`.
Step 3: Return `True` if both checks are either `True` or both are `False`. Return `False` otherwise.

Key Insights

Insight 1: The crucial insight is recognizing that the presence or absence of '1's in the string is invariant under the given operation. The total number of '1's may change, but if there's at least one '1' initially, the operation can propagate it. If there are no '1's initially, the operation cannot create one.
Insight 2: The `OR` and `XOR` operations, when applied together in the specified manner, ensure that the total number of bits set to '1' are either preserved or potentially increased. The presence of even one '1' in the string allows us to potentially set any other bit to '1' via the described operation.
Insight 3: Therefore, the core requirement is to check if both `s` and `target` either both contain at least one '1', or both contain only '0's. This is sufficient to determine whether the transformation is possible.

Complexity Analysis

Time Complexity: O(n), where n is the length of the string. The `in` operator in Python has a time complexity of O(n) in the worst case, as it might need to scan the entire string.

Space Complexity: O(1). The solution uses a constant amount of extra space, regardless of the input size.

Topics

This problem involves: String, Bit Manipulation.

Companies

Asked at: Sprinklr.

Apply Bitwise Operations to Make Strings Equal | LeetCode 2546

Medium

String Bit Manipulation

Time: O(n), where n is the length of the string. The `in` operator in Python has a time complexity of O(n) in the worst case, as it might need to scan the entire string.Space: O(1). The solution uses a constant amount of extra space, regardless of the input size.

Solution Code

Available in:

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2546: Apply Bitwise Operations to Make Strings Equal - Detailed explanation and solution approaches.

Topics

String Bit Manipulation

Companies

Sprinklr

Stats

Difficulty:Medium

Languages:2

Companies:1

Problem #:2546

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

3 steps

1
Step 1: Check if the character '1' is present in string `s`.
2
Step 2: Check if the character '1' is present in string `target`.
3
Step 3: Return `True` if both checks are either `True` or both are `False`. Return `False` otherwise.

💡

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(n), where n is the length of the string. The `in` operator in Python has a time complexity of O(n) in the worst case, as it might need to scan the entire string.

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1). The solution uses a constant amount of extra space, regardless of the input size.

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the `in` operator which iterates through the strings `s` and `target` at most once each. The space complexity is constant because we are only using boolean values to store intermediate results.

📖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

Counting Ones

Instead of using the `in` operator, you can iterate through the strings and count the number of '1's in each. Then compare if both counts are either zero or non-zero.

Time: O(n), where n is the length of the strings.Space: O(1)

Trade-offs: This approach is functionally equivalent to the provided solution in terms of performance, but might be slightly more verbose. It explicitly counts the '1's instead of relying on the implicit check by 'in'.

✨

Key Insights

1Insight 1: The crucial insight is recognizing that the presence or absence of '1's in the string is invariant under the given operation. The total number of '1's may change, but if there's at least one '1' initially, the operation can propagate it. If there are no '1's initially, the operation cannot create one.
2Insight 2: The `OR` and `XOR` operations, when applied together in the specified manner, ensure that the total number of bits set to '1' are either preserved or potentially increased. The presence of even one '1' in the string allows us to potentially set any other bit to '1' via the described operation.
3Insight 3: Therefore, the core requirement is to check if both `s` and `target` either both contain at least one '1', or both contain only '0's. This is sufficient to determine whether the transformation is possible.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: `s = '0000', target = '1111'`. The solution correctly returns `False` because `s` has no '1's but `target` does.
Edge case 2: `s = '1000', target = '0000'`. The solution correctly returns `False` because `s` has '1' but `target` does not.
Edge case 3: `s = '1010', target = '1111'`. The solution correctly returns `True` because both strings contain at least one '1'.
Edge case 4: `s = '0000', target = '0000'`. The solution correctly returns `True` because neither string contains a '1'.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Thinking that the number of '1's must be the same. The total number of '1's may be different, it only matters whether there's at least one.

✗Common mistake 2: Trying to simulate the bitwise operations. This is unnecessary and would lead to a much more complex and less efficient solution. Recognizing the invariant is key.

✗Common mistake 3: Forgetting the case where both strings are all zeros. If you only focus on the presence of 1s in s to make target, you might wrongly return False when both strings have no 1s.

✅

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: Error correction codes. In some coding schemes, the presence or absence of a parity bit (a single bit indicating whether the number of 1s is even or odd) determines the validity of the transmitted data. The principle of preserving the existence of a '1' is analogous to checking parity.

🏦Real-world scenario 2: Data integrity checks. If a system depends on at least one bit being set in a specific data field, this approach can be used to quickly verify that the integrity of that field has been maintained after some transformation or transmission.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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: June 27, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

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.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Apply Bitwise Operations to Make Strings Equal"?

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

🤖 Need help with this problem?