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

This problem uses String, Dynamic Programming, Sliding Window. Understanding these concepts is essential for solving this Medium problem efficiently.

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

Insight 1: Rotating the string allows us to consider all possible starting positions without actually performing the rotations. This is achieved by effectively concatenating the string with itself. Insight 2: There are only two possible alternating patterns: starting with '0' or starting with '1'. We need to check both.

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Minimum Number of Flips to Make the Binary String Alternating - LeetCode 1888 Solution

Q: Which companies ask LeetCode 1888. Minimum Number of Flips to Make the Binary String Alternating in interviews?

This problem is frequently asked at IBM.

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

Minimum Number of Flips to Make the Binary String Alternating is LeetCode problem 1888, 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

You are given a binary string s . You are allowed to perform two types of operations on the string in any sequence: Type-1: Remove the character at the start of the string s and append it to the end of the string. Type-2: Pick any character in s and flip its value, i.e., if its value is '0' it becomes '1' and vice-versa. Return the minimum number of type-2 operations you need to perform such that s becomes alternating . The string is called alternating if no two adjacent characters are equal. Fo

Detailed Explanation

The problem asks us to find the minimum number of flips (changing 0 to 1 or 1 to 0) needed to make a binary string alternating. We can also rotate the string by moving the first character to the end. The goal is to find the fewest flips required across all possible rotations to achieve an alternating pattern (e.g., '010101' or '101010').

Solution Approach

The solution utilizes a sliding window technique. It first doubles the input string to simulate rotations. Then, it calculates the number of mismatches (flips needed) with respect to the two possible alternating patterns ('0101...' and '1010...') for the initial window of length 'n'. The sliding window then moves one position at a time, updating the mismatch counts by subtracting the mismatch count of the character leaving the window and adding the mismatch count of the character entering the window. The minimum number of flips across all window positions is then returned.

Step-by-Step Algorithm

Step 1: Double the input string `s` to create `s_double`. This helps simulate rotations without explicitly performing them.
Step 2: Create two arrays, `mismatches_p0` and `mismatches_p1`, to store the number of mismatches with the '0101...' pattern (P0) and the '1010...' pattern (P1), respectively, for each character in `s_double`.
Step 3: Iterate through `s_double` and calculate the values for `mismatches_p0` and `mismatches_p1`. If a character at index `i` doesn't match the expected character in the corresponding alternating pattern, increment the mismatch count.
Step 4: Calculate the initial number of flips needed for the first window (length `n`) for both patterns. This involves summing the mismatch counts from `mismatches_p0` and `mismatches_p1` for the first `n` characters.
Step 5: Initialize `min_flips` to the minimum of the initial `flips_p0` and `flips_p1`.
Step 6: Iterate through the remaining windows (simulating rotations). For each window, update `flips_p0` and `flips_p1` by subtracting the mismatch count of the character leaving the window and adding the mismatch count of the character entering the window.
Step 7: Update `min_flips` with the minimum value found so far between `min_flips`, `flips_p0`, and `flips_p1`.
Step 8: Return `min_flips`.

Key Insights

Insight 1: Rotating the string allows us to consider all possible starting positions without actually performing the rotations. This is achieved by effectively concatenating the string with itself.
Insight 2: There are only two possible alternating patterns: starting with '0' or starting with '1'. We need to check both.
Insight 3: Using a sliding window approach significantly reduces the computational effort by only updating the flip counts rather than recalculating them for each rotation.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Dynamic Programming, Sliding Window.

Companies

Asked at: IBM.

Minimum Number of Flips to Make the Binary String Alternating | LeetCode 1888

Medium

String Dynamic Programming Sliding Window

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

Solution Code

Available in:

class Solution {
    public int minFlips(String s) {
        int n = s.length();
        String s_double = s + s;
        
        // Mismatches against two global alternating patterns of length 2n
        // P0 = "0101..."
        // P1 = "1010..."
        
        int[] mismatches_p0 = new int[2 * n];
        int[] mismatches_p1 = new int[2 * n];
        
        for (int i = 0; i < 2 * n; i++) {
            if (i % 2 == 0) {
                if (s_double.charAt(i) == '1') {
                    mismatches_p0[i] = 1;
                } else { // s_double.charAt(i) == '0'
                    mismatches_p1[i] = 1;
                }
            } else { // i % 2 == 1
                if (s_double.charAt(i) == '0') {
                    mismatches_p0[i] = 1;
                } else { // s_double.charAt(i) == '1'
                    mismatches_p1[i] = 1;
                }
            }
        }
        
        // Use a sliding window of size n to check all rotations of s
        int flips_p0 = 0;
        for (int i = 0; i < n; i++) {
            flips_p0 += mismatches_p0[i];
        }
        
        int flips_p1 = 0;
        for (int i = 0; i < n; i++) {
            flips_p1 += mismatches_p1[i];
        }
        
        int min_flips = Integer.MAX_VALUE;
        
        // For a window starting at index i, s_double[i:i+n],
        // we calculate the flips needed to make it alternating.
        // An alternating string must match either "0101..." or "1010...".
        // If i is even, the cost to match "0101..." is flips_p0.
        // If i is odd, the cost to match "0101..." is flips_p1.
        // Regardless, the min flips for the window is min(flips_p0, flips_p1).
        
        min_flips = Math.min(flips_p0, flips_p1);
        
        for (int i = 1; i < n; i++) {
            // Update window sums
            flips_p0 = flips_p0 - mismatches_p0[i-1] + mismatches_p0[i+n-1];
            flips_p1 = flips_p1 - mismatches_p1[i-1] + mismatches_p1[i+n-1];
            
            min_flips = Math.min(min_flips, Math.min(flips_p0, flips_p1));
        }
                
        return min_flips;
    }
}

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Problem Info

Description

Problem 1888: Minimum Number of Flips to Make the Binary String Alternating - Detailed explanation and solution approaches.

Topics

String Dynamic Programming Sliding Window

Companies

IBM

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1888

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

8 steps

1
Step 1: Double the input string `s` to create `s_double`. This helps simulate rotations without explicitly performing them.
2
Step 2: Create two arrays, `mismatches_p0` and `mismatches_p1`, to store the number of mismatches with the '0101...' pattern (P0) and the '1010...' pattern (P1), respectively, for each character in `s_double`.
3
Step 3: Iterate through `s_double` and calculate the values for `mismatches_p0` and `mismatches_p1`. If a character at index `i` doesn't match the expected character in the corresponding alternating pattern, increment the mismatch count.
4
Step 4: Calculate the initial number of flips needed for the first window (length `n`) for both patterns. This involves summing the mismatch counts from `mismatches_p0` and `mismatches_p1` for the first `n` characters.
5
Step 5: Initialize `min_flips` to the minimum of the initial `flips_p0` and `flips_p1`.
6
Step 6: Iterate through the remaining windows (simulating rotations). For each window, update `flips_p0` and `flips_p1` by subtracting the mismatch count of the character leaving the window and adding the mismatch count of the character entering the window.
7
Step 7: Update `min_flips` with the minimum value found so far between `min_flips`, `flips_p0`, and `flips_p1`.
8
Step 8: Return `min_flips`.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the loops that iterate through the doubled string. Since the length of the doubled string is proportional to the length of the original string, the time complexity is linear, O(n). The space complexity arises primarily from the creation of the doubled string and the two mismatch arrays, all of which are linearly proportional to the input string's length, hence O(n).

📖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

Dynamic Programming

Instead of a sliding window, a dynamic programming approach could be used to calculate the minimum flips. However, it would be more complex to implement and wouldn't necessarily offer better performance.

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

Trade-offs: The sliding window approach is generally preferred due to its simplicity and efficient implementation. The dynamic programming approach may provide no significant advantage and can be harder to understand.

✨

Key Insights

1Insight 1: Rotating the string allows us to consider all possible starting positions without actually performing the rotations. This is achieved by effectively concatenating the string with itself.
2Insight 2: There are only two possible alternating patterns: starting with '0' or starting with '1'. We need to check both.
3Insight 3: Using a sliding window approach significantly reduces the computational effort by only updating the flip counts rather than recalculating them for each rotation.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Sliding Window

A moving "window" over data to process subsets efficiently.

String

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

Edge Cases

Edge case 1: Input string `s` is already alternating (e.g., '0101'). The algorithm correctly returns 0 because no flips are needed.
Edge case 2: Input string `s` consists of all the same characters (e.g., '0000' or '1111'). The algorithm correctly determines the minimum flips needed to make it alternating (which would be n/2, rounded down, if n is even and (n+1)/2, rounded down, if n is odd).
Edge case 3: Input string `s` has length 1. The algorithm correctly returns 0, because a string of length 1 is always considered alternating.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to consider both alternating patterns ('0101...' and '1010...') and only focusing on one.

✗Common mistake 2: Failing to account for rotations of the string and only considering the initial string without rotating.

✗Common mistake 3: Inefficiently recalculating the number of flips for each rotation instead of using the sliding window technique to update the count.

✅

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. Alternating patterns are easier to compress, so finding the minimum flips needed to achieve such a pattern can optimize compression algorithms.

🏦Real-world scenario 2: Error correction in data transmission. In scenarios where data is transmitted as binary strings, ensuring alternating patterns can aid in detecting and correcting errors introduced during 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

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 28, 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 13

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

@TechLeadJul 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 identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

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

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