What is the time complexity of Minimum Number of Swaps to Make the Binary String Alternating?

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

What data structures are used to solve Minimum Number of Swaps to Make the Binary String Alternating?

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

What are the key insights for solving Minimum Number of Swaps to Make the Binary String Alternating?

Insight 1: The number of '0's and '1's in the target alternating string can differ by at most 1. If the difference is greater than 1, an alternating string cannot be formed, and -1 should be returned. Insight 2: If the length of the string is odd, the alternating string is uniquely determined by whether '1' or '0' appears more frequently. The alternating string must start with the more frequent character.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Minimum Number of Swaps to Make the Binary String Alternating - LeetCode 1864 Solution

Q: Which companies ask LeetCode 1864. Minimum Number of Swaps to Make the Binary String Alternating in interviews?

This problem is frequently asked at Societe Generale.

Minimum Number of Swaps to Make the Binary String Alternating - Complete Solution Guide

Minimum Number of Swaps to Make the Binary String Alternating is LeetCode problem 1864, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a binary string s , return the minimum number of character swaps to make it alternating , or -1 if it is impossible. The string is called alternating if no two adjacent characters are equal. For example, the strings "010" and "1010" are alternating, while the string "0100" is not. Any two characters may be swapped, even if they are not adjacent . Example 1: Input: s = "111000" Output: 1 Explanation: Swap positions 1 and 4: "1 1 10 0 0" -> "1 0 10 1 0" The string is now alternating. Example

Detailed Explanation

The problem requires finding the minimum number of swaps needed to transform a given binary string into an alternating binary string (e.g., "01010" or "10101"). An alternating string has no two adjacent characters that are the same. If it's impossible to create an alternating string from the given input (due to an imbalance in the number of '0's and '1's), the function should return -1. Swaps can be performed between any two characters in the string, regardless of their position.

Solution Approach

The solution first checks if forming an alternating string is possible by verifying that the absolute difference between the count of '0's and '1's is no more than 1. If it is possible, the solution proceeds based on whether the string length is odd or even. If odd, there is only one possible alternating pattern. If even, there are two possible alternating patterns, and the solution computes the swaps for each and returns the minimum. The core idea is that the minimum swaps required is the number of digits placed incorrectly in the 'even' indices of the candidate alternating pattern.

Step-by-Step Algorithm

Step 1: Count the number of '1's and '0's in the given string.
Step 2: Check if the absolute difference between the counts of '1's and '0's is greater than 1. If so, return -1 (impossible to form an alternating string).
Step 3: If the length of the string is odd:
Step 3a: Determine whether '1's or '0's are more frequent. The alternating string must start with the more frequent character.
Step 3b: Iterate through the string, considering only characters at even indices (0, 2, 4, ...).
Step 3c: Count the number of misplaced characters at even indices. If the target string starts with '1', count '0's. If the target string starts with '0', count '1's.
Step 3d: Return the misplaced count as the minimum number of swaps.
Step 4: If the length of the string is even:
Step 4a: Calculate the number of swaps needed to make the string alternating starting with '0' (count misplaced '1's at even indices).
Step 4b: Calculate the number of swaps needed to make the string alternating starting with '1' (count misplaced '0's at even indices).
Step 4c: Return the minimum of the two swap counts calculated above.

Key Insights

Insight 1: The number of '0's and '1's in the target alternating string can differ by at most 1. If the difference is greater than 1, an alternating string cannot be formed, and -1 should be returned.
Insight 2: If the length of the string is odd, the alternating string is uniquely determined by whether '1' or '0' appears more frequently. The alternating string must start with the more frequent character.
Insight 3: If the length of the string is even, there are two possible alternating strings: one starting with '0' and the other starting with '1'. We need to calculate the minimum swaps needed for both cases and return the smaller value.
Insight 4: Counting the misplaced characters in one of the alternating string patterns gives the minimum number of swaps needed.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Greedy.

Companies

Asked at: Societe Generale.

Minimum Number of Swaps to Make the Binary String Alternating | LeetCode 1864

Medium

String Greedy

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

Solution Code

Available in:

class Solution {
    public int minSwaps(String s) {
        int n = s.length();
        int ones = 0;
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == '1') {
                ones++;
            }
        }
        int zeros = n - ones;

        if (Math.abs(ones - zeros) > 1) {
            return -1;
        }

        if (n % 2 == 1) {
            if (ones > zeros) {
                // Target must start with '1': "101..."
                // Count misplaced '0's, which are at even positions.
                // Each misplaced '0' at an even position needs one swap.
                int misplaced = 0;
                for (int i = 0; i < n; i += 2) {
                    if (s.charAt(i) == '0') {
                        misplaced++;
                    }
                }
                return misplaced;
            } else {
                // Target must start with '0': "010..."
                // Count misplaced '1's, which are at even positions.
                int misplaced = 0;
                for (int i = 0; i < n; i += 2) {
                    if (s.charAt(i) == '1') {
                        misplaced++;
                    }
                }
                return misplaced;
            }
        } else {
            // Even length, two possible targets.
            // Target 1: "0101..."
            // Swaps needed is the number of '1's at even positions.
            int swaps1 = 0;
            for (int i = 0; i < n; i += 2) {
                if (s.charAt(i) == '1') {
                    swaps1++;
                }
            }

            // Target 2: "1010..."
            // Swaps needed is the number of '0's at even positions.
            int swaps2 = 0;
            for (int i = 0; i < n; i += 2) {
                if (s.charAt(i) == '0') {
                    swaps2++;
                }
            }

            return Math.min(swaps1, swaps2);
        }
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1864: Minimum Number of Swaps to Make the Binary String Alternating - Detailed explanation and solution approaches.

Topics

String Greedy

Companies

Societe Generale

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1864

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: Count the number of '1's and '0's in the given string.
2
Step 2: Check if the absolute difference between the counts of '1's and '0's is greater than 1. If so, return -1 (impossible to form an alternating string).
3
Step 3: If the length of the string is odd:
4
Step 3a: Determine whether '1's or '0's are more frequent. The alternating string must start with the more frequent character.
5
Step 3b: Iterate through the string, considering only characters at even indices (0, 2, 4, ...).
6
Step 3c: Count the number of misplaced characters at even indices. If the target string starts with '1', count '0's. If the target string starts with '0', count '1's.
7
Step 3d: Return the misplaced count as the minimum number of swaps.
8
Step 4: If the length of the string is even:
9
Step 4a: Calculate the number of swaps needed to make the string alternating starting with '0' (count misplaced '1's at even indices).
10
Step 4b: Calculate the number of swaps needed to make the string alternating starting with '1' (count misplaced '0's at even indices).
11
Step 4c: Return the minimum of the two swap counts calculated above.

💡

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 time complexity is determined by the number of iterations through the string, which is linear to the length of the string. The space complexity is constant because the solution only utilizes a fixed amount of extra memory to store the counts, misplaced characters, and the string length.

📖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 XOR and Prefix Sum

We can calculate the number of swaps by keeping track of expected values at each index of the alternating string and comparing them with actual values using XOR. Use prefix sums to keep track of number of mismatches.

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

Trade-offs: This approach is slightly harder to understand but achieves the same time and space complexity. May be less readable for beginners.

✨

Key Insights

1Insight 1: The number of '0's and '1's in the target alternating string can differ by at most 1. If the difference is greater than 1, an alternating string cannot be formed, and -1 should be returned.
2Insight 2: If the length of the string is odd, the alternating string is uniquely determined by whether '1' or '0' appears more frequently. The alternating string must start with the more frequent character.
3Insight 3: If the length of the string is even, there are two possible alternating strings: one starting with '0' and the other starting with '1'. We need to calculate the minimum swaps needed for both cases and return the smaller value.
4Insight 4: Counting the misplaced characters in one of the alternating string patterns gives the minimum number of swaps needed.

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

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

Edge Cases

Edge case 1: Empty string. The code handles this correctly implicitly because the loops won't execute and the counts will be zero, and `abs(ones - zeros)` will be zero.
Edge case 2: String already alternating. The algorithm correctly returns 0 in this case as there will be no misplaced characters.
Edge case 3: String with only 0s or only 1s. The algorithm will correctly calculate the number of swaps needed, or return -1 if the difference between number of zeros and ones is greater than one.
Edge case 4: String with an imbalance in the counts of zeros and ones that make alternating impossible. The code correctly returns -1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking the difference between the number of '0's and '1's. Forgetting to verify if a solution is even possible leads to incorrect calculations.

✗Common mistake 2: Incorrectly calculating the number of swaps when the string length is even. Forgetting to consider both alternating patterns (starting with '0' and starting with '1') and picking the minimum is a common error.

✗Common mistake 3: Assuming that all '0's need to be moved to even positions when the string is odd. It depends on whether '0' or '1' occurs more frequently. Misplacing the '0's or '1's only at even indices when computing swaps.

✅

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: Data compression. Algorithms for data compression might use alternating patterns to represent data, and this algorithm could be used to minimize the number of transformations needed to achieve the desired compressed representation.

🏦Real-world scenario 2: Error correction. In communication systems, alternating patterns are sometimes used as synchronization signals. This algorithm could be useful in correcting errors in received signals by minimizing the number of bit flips needed to restore the expected pattern.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: August 4, 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

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 do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

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 "Minimum Number of Swaps to Make the Binary String Alternating"?

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 greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?