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

Master LeetCode problem 1888 (Minimum Number of Flips to Make the Binary String Alternating), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1888. Minimum Number of Flips to Make the Binary String Alternating

Medium

String Dynamic Programming Greedy Sliding Window

Problem Explanation

Explanation

To solve this problem, we can simulate two scenarios: one where we start with '0' as the first character and the other where we start with '1' as the first character. We calculate the number of flips needed for both scenarios and return the minimum flips required.

We iterate over the string and count the number of flips needed to make the current character match the expected character ('0' or '1' depending on the starting character).
For each scenario, we calculate the total flips needed by summing up the flips required to make the string alternating.
Finally, we return the minimum of the two scenarios as the answer.

Solution Code

class Solution {
    public int minFlips(String s) {
        int n = s.length();
        s += s; // Concatenate to handle circular property

        int minFlipsStartWithZero = 0;
        int minFlipsStartWithOne = 0;

        for (int i = 0; i < 2 * n; i++) {
            char expectedZero = (i % 2 == 0) ? '0' : '1';
            char expectedOne = (i % 2 == 0) ? '1' : '0';

            if (s.charAt(i) != expectedZero) {
                minFlipsStartWithZero++;
            }

            if (s.charAt(i) != expectedOne) {
                minFlipsStartWithOne++;
            }

            if (i >= n) {
                char toRemove = (s.charAt(i - n) == '0') ? '0' : '1';
                minFlipsStartWithZero -= (s.charAt(i - n) != expectedZero) ? 1 : 0;
                minFlipsStartWithOne -= (s.charAt(i - n) != expectedOne) ? 1 : 0;
            }
        }

        return Math.min(minFlipsStartWithZero, minFlipsStartWithOne);
    }
}

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