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Leetcode 1758. Minimum Changes To Make Alternating Binary String

Problem Statement

Given a binary string s, you need to find the minimum number of character flips required to make the string alternating. An alternating binary string is a string where no two adjacent characters are the same. For example, “010101” and “101010” are alternating binary strings. You need to return the minimum number of flips required.

Clarifying Questions

  1. What’s the length of the string?
    • The length can vary; it’s not specified in the problem, but it should be within reasonable limits typically seen in coding challenges.
  2. What characters does the string contain?
    • The string contains only characters ‘0’ and ‘1’.
  3. Can the string be empty?
    • According to typical constraints, the string should have at least one character.

Strategy

To solve this problem, consider these two possible alternating patterns:

  1. “010101…” (starting with ‘0’)
  2. “101010…” (starting with ‘1’)

For each pattern, we count the number of positions where the actual character in the string s does not match the expected character in the pattern. The minimum count between these two results will be the minimum number of flips required to make the string alternating.

Code

Here is the code to implement the solution in Java:

public class MinimumFlipsToAlternatingString {
    public int minFlips(String s) {
        int n = s.length();
        int flips1 = 0; // For pattern '010101...'
        int flips2 = 0; // For pattern '101010...'
        
        for (int i = 0; i < n; i++) {
            char expectedCharPattern1 = (i % 2 == 0) ? '0' : '1';
            char expectedCharPattern2 = (i % 2 == 0) ? '1' : '0';
            
            if (s.charAt(i) != expectedCharPattern1) flips1++;
            if (s.charAt(i) != expectedCharPattern2) flips2++;
        }
        
        return Math.min(flips1, flips2);
    }
    
    public static void main(String[] args) {
        MinimumFlipsToAlternatingString solution = new MinimumFlipsToAlternatingString();
        
        System.out.println(solution.minFlips("01001001101")); // Example test case
    }
}

Time Complexity

The time complexity of this solution is O(n), where n is the length of the string. This is because we iterate through each character of the string once to evaluate whether it matches the expected character in both potential alternating patterns.

Explanation

  1. Initialize two counters: flips1 and flips2 to count mismatches for both patterns.
  2. Loop through each character in the string:
    • Determine the expected character for both patterns at the current index.
    • Increment the respective counter if the character does not match the pattern’s expected character.
  3. Return the minimum of the two counters: which gives the minimum number of flips required.

I hope this helps! Let me know if you have any questions.

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