What is the time complexity of Minimum Changes To Make Alternating Binary String?

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

What data structures are used to solve Minimum Changes To Make Alternating Binary String?

This problem uses String. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Minimum Changes To Make Alternating Binary String?

Insight 1: There are only two possible alternating patterns: starting with '0' ("0101...") and starting with '1' ("1010..."). Insight 2: We can count the number of changes needed for each pattern and choose the smaller one.

Which companies ask LeetCode 1758. Minimum Changes To Make Alternating Binary String in interviews?

This problem is frequently asked at Tesla.

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Minimum Changes To Make Alternating Binary String - LeetCode 1758 Solution

Minimum Changes To Make Alternating Binary String - Complete Solution Guide

Minimum Changes To Make Alternating Binary String is LeetCode problem 1758, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a string s consisting only of the characters '0' and '1' . In one operation, you can change any '0' to '1' or vice versa. The string is called alternating if no two adjacent characters are equal. For example, the string "010" is alternating, while the string "0100" is not. Return the minimum number of operations needed to make s alternating . Example 1: Input: s = "0100" Output: 1 Explanation: If you change the last character to '1', s will be "0101", which is alternating. Example

Detailed Explanation

The problem asks you to find the minimum number of changes needed to transform a binary string (containing only '0's and '1's) into an alternating binary string. An alternating string is defined as a string where no two adjacent characters are the same. For example, "01010" is alternating, while "0011" is not. The only operation allowed is changing a '0' to a '1' or vice-versa. The input is a binary string, and the output is an integer representing the minimum number of changes required.

Solution Approach

The solution uses a greedy approach. It iterates through the input string and counts the number of changes needed to create two alternating patterns: one starting with '0' and another starting with '1'. It then returns the minimum of these two counts. This approach works because for any given string, transforming it to an alternating pattern will always involve changing some subset of its characters. By checking both patterns, we guarantee finding the optimal minimum.

Step-by-Step Algorithm

Step 1: Initialize two counters, `count1` and `count2`, both to 0. `count1` will track the changes needed to create an alternating string starting with '0', and `count2` will track changes for an alternating string starting with '1'.
Step 2: Iterate through the input string `s`. For each character:
Step 3: If the index `i` is even:
Step 4: If the character `s[i]` is '1', increment `count1` (because we'd need to change it to '0' to match the '0101...' pattern). Otherwise, increment `count2` (because we need to change '0' to '1' to match '1010...').
Step 5: If the index `i` is odd:
Step 6: If the character `s[i]` is '0', increment `count1` (to change to '1'). Otherwise, increment `count2` (to change to '0').
Step 7: After iterating through the entire string, return the minimum of `count1` and `count2`.

Key Insights

Insight 1: There are only two possible alternating patterns: starting with '0' ("0101...") and starting with '1' ("1010...").
Insight 2: We can count the number of changes needed for each pattern and choose the smaller one.
Insight 3: The problem can be solved efficiently in a single pass through the string without the need for recursion or complex data structures.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String.

Companies

Asked at: Tesla.

Minimum Changes To Make Alternating Binary String | LeetCode 1758

Easy

String

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minOperations(String s) {
        int n = s.length();
        int count1 = 0;
        int count2 = 0;
        for (int i = 0; i < n; i++) {
            if (i % 2 == 0) {
                if (s.charAt(i) == '1') {
                    count1++;
                } else {
                    count2++;
                }
            } else {
                if (s.charAt(i) == '0') {
                    count1++;
                } else {
                    count2++;
                }
            }
        }
        return Math.min(count1, count2);
    }
}

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Problem Info

Description

Problem 1758: Minimum Changes To Make Alternating Binary String - Detailed explanation and solution approaches.

Topics

String

Companies

Tesla

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1758

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

7 steps

1
Step 1: Initialize two counters, `count1` and `count2`, both to 0. `count1` will track the changes needed to create an alternating string starting with '0', and `count2` will track changes for an alternating string starting with '1'.
2
Step 2: Iterate through the input string `s`. For each character:
3
Step 3: If the index `i` is even:
4
Step 4: If the character `s[i]` is '1', increment `count1` (because we'd need to change it to '0' to match the '0101...' pattern). Otherwise, increment `count2` (because we need to change '0' to '1' to match '1010...').
5
Step 5: If the index `i` is odd:
6
Step 6: If the character `s[i]` is '0', increment `count1` (to change to '1'). Otherwise, increment `count2` (to change to '0').
7
Step 7: After iterating through the entire string, return the minimum of `count1` and `count2`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates through the input string once, performing constant-time operations in each iteration. Therefore, the time complexity is linear with respect to the length of the string (n). The algorithm uses only a few integer variables to store counts; the space used is constant regardless of the input size.

📖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 Bit Manipulation

Instead of explicit character comparison, bit manipulation could be used to check and increment counters. This might offer slight performance improvement in some environments.

Time: O(n)Space: O(1)

Trade-offs: Bit manipulation can be slightly faster in some contexts, but it might make the code less readable. The provided approach is clearer and easier to understand for most programmers.

✨

Key Insights

1Insight 1: There are only two possible alternating patterns: starting with '0' ("0101...") and starting with '1' ("1010...").
2Insight 2: We can count the number of changes needed for each pattern and choose the smaller one.
3Insight 3: The problem can be solved efficiently in a single pass through the string without the need for recursion or complex data structures.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: Empty string. The code handles this correctly; the loop doesn't run, and `min(count1, count2)` will return 0, which is the correct answer.
Edge case 2: String with all 0s or all 1s. The code correctly computes half the string length as the minimum number of operations needed.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to consider both alternating patterns ('0101...' and '1010...'). A solution that only considers one pattern will not always find the minimum number of operations.

✗Common mistake 2: Off-by-one errors in the loop index or conditionals. Careful attention to index parity (even/odd) is crucial for this problem.

✅

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: Optimizing data transmission by minimizing bit flips in a communication system. An alternating bit pattern might be preferred for error detection or synchronization.

🏦Real-world scenario 2: Designing efficient encoding schemes where minimizing changes in a sequence can reduce power consumption or improve transmission speed.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: October 17, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

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

@TechLeadJul 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.

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 "Minimum Changes To Make Alternating Binary String"?

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

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