Minimum Number of Flips to Make the Binary String Alternating — Algorithm Visualization & Coding Challenge

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Understanding Minimum Number of Flips to Make the Binary String Alternating

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Problem Overview

🔄 Minimum Number of Flips to Make the Binary String Alternating

Difficulty: 🟡 Medium
Topics: Greedy, Sliding Window, String

🧠 Problem Statement

You are given a binary string s.

You are allowed to perform two types of operations in any sequence:

🔁 Type-1 Operation (Rotation)

Remove the character at the start of the string and append it to the end.

🔄 Type-2 Operation (Flip)

Pick any character in s and flip its value:

'0' → '1'
'1' → '0'

🎯 Goal

Return the minimum number of Type-2 (flip) operations required to make the string alternating.

A string is called alternating if no two adjacent characters are equal.

✔️ Valid alternating strings:

"010"
"1010"

❌ Not alternating:

"0100"

📌 Examples

Example 1

Input:  s = "111000"
Output: 2

Explanation:

Perform Type-1 operation twice → "100011"
Flip 3rd and 6th characters → "101010"
Total flips = 2

Example 2

Input:  s = "010"
Output: 0

Explanation: The string is already alternating ✅

Example 3

Input:  s = "1110"
Output: 1

Explanation:

Flip 2nd character → "1010"
Total flips = 1

🔒 Constraints

1 <= s.length <= 10^5
s[i] is either '0' or '1'

💡 Key Insight

Since rotation is allowed (Type-1), we must consider all cyclic shifts of the string.
For each possible rotation, compute the minimum flips required to match:

Pattern 1 → 010101...
Pattern 2 → 101010...

Return the minimum among all possibilities.

💻 Java Code

java

public class Main {

    public static void main(String[] args) {

        // 🔹 Hardcoded Input
        String s = "111000";

        int result = minFlips(s);

        System.out.println("Input: " + s);
        System.out.println("Minimum Flips Required: " + result);
    }

    public static int minFlips(String s) {

        int n = s.length();

        // 🔹 Double the string to simulate rotations
        String doubled = s + s;

        // 🔹 Create alternating patterns of size 2n
        StringBuilder alt1 = new StringBuilder(); // 010101...
        StringBuilder alt2 = new StringBuilder(); // 101010...

        for (int i = 0; i < 2 * n; i++) {
            if (i % 2 == 0) {
                alt1.append('0');
                alt2.append('1');
            } else {
                alt1.append('1');
                alt2.append('0');
            }
        }

        int left = 0;
        int diff1 = 0;
        int diff2 = 0;
        int answer = Integer.MAX_VALUE;

        // 🔹 Sliding window
        for (int right = 0; right < 2 * n; right++) {

            if (doubled.charAt(right) != alt1.charAt(right))
                diff1++;

            if (doubled.charAt(right) != alt2.charAt(right))
                diff2++;

            // Maintain window size = n
            if (right - left + 1 > n) {

                if (doubled.charAt(left) != alt1.charAt(left))
                    diff1--;

                if (doubled.charAt(left) != alt2.charAt(left))
                    diff2--;

                left++;
            }

            if (right - left + 1 == n) {
                answer = Math.min(answer, Math.min(diff1, diff2));
            }
        }

        return answer;
    }
}

⏱ Complexity

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

✨ Perfect problem to practice Sliding Window + Alternating Pattern Matching

— Written by Saurabh Patil • B.Tech CSE • Software Developer

Reference Link

https://leetcode.com/problems/minimum-number-of-flips-to-make-the-binary-string-alternating/description/

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