LeetCode 3445: Maximum Difference Between Even and Odd Frequency II Solution

Master LeetCode problem 3445 (Maximum Difference Between Even and Odd Frequency II), a hard 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.

3445. Maximum Difference Between Even and Odd Frequency II

Hard

String Sliding Window Enumeration Prefix Sum

Problem Explanation

Explanation:

To solve this problem, we can iterate through all substrings of length at least k and keep track of the frequencies of characters within each substring. For each substring, we calculate the difference between the frequency of the character with odd frequency and the frequency of the character with even frequency. We update the maximum difference found so far. Finally, we return the maximum difference.

Algorithm:

Initialize a variable maxDiff to store the maximum difference found so far.
Iterate through all substrings of length at least k in the input string s.
For each substring, calculate the frequencies of all characters.
Find the character with odd frequency and the character with even frequency in the substring and calculate the difference in their frequencies.
Update maxDiff if the calculated difference is greater than the current maxDiff.
Return the final maxDiff as the result.

Time Complexity: O(N^2) where N is the length of the input string s. Space Complexity: O(N) for storing frequencies of characters.

Solution Code

class Solution {
    public int maxDiff(String s, int k) {
        int maxDiff = -1;
        for (int i = 0; i <= s.length() - k; i++) {
            int[] freq = new int[5];
            for (int j = i; j < i + k; j++) {
                freq[s.charAt(j) - '0']++;
            }
            for (int a = 0; a <= 4; a++) {
                for (int b = 0; b <= 4; b++) {
                    if (freq[a] > 0 && freq[b] > 0 && freq[a] % 2 == 1 && freq[b] % 2 == 0) {
                        maxDiff = Math.max(maxDiff, freq[a] - freq[b]);
                    }
                }
            }
        }
        return maxDiff;
    }
}

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