What is the brute force solution for Maximum Difference Between Even and Odd Frequency II?

The brute force approach for Maximum Difference Between Even and Odd Frequency II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible substrings of the given string to find the maximum difference between the frequencies of two characters. This method is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Difference Between Even and Odd Frequency II?

Maximum Difference Between Even and Odd Frequency II uses the following concepts: String, Sliding Window, Enumeration, Prefix Sum. The recommended patterns to study are: Sliding Window, Frequency Count.

What are the interview tips for Maximum Difference Between Even and Odd Frequency II?

Always clarify the problem constraints and requirements before jumping into coding. Think about edge cases, such as strings with limited character diversity. Practice explaining your thought process as you code; it helps in interviews.

What are common mistakes when solving Maximum Difference Between Even and Odd Frequency II?

Failing to check both odd and even frequencies before calculating the difference. Not maintaining the window size correctly, which can lead to incorrect frequency counts.

#3445

Maximum Difference Between Even and Odd Frequency II

Hard

StringSliding WindowEnumerationPrefix SumSliding WindowFrequency Count

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a sliding window technique combined with character frequency tracking. By maintaining a window of characters, we can efficiently calculate the required frequencies and differences without generating all substrings.

⚙️

Algorithm

4 steps

1Step 1: Initialize a frequency array to count occurrences of characters in the current window.
2Step 2: Use two pointers to define the window, expanding the right pointer to include more characters.
3Step 3: Once the window size is at least k, calculate the maximum odd frequency and minimum even frequency.
4Step 4: Move the left pointer to shrink the window while maintaining the condition of at least k characters, updating frequencies accordingly.

solution.py16 lines

1def maxDifference(s, k):
2    freq = [0] * 10
3    max_diff = float('-inf')
4    left = 0
5    for right in range(len(s)):
6        freq[int(s[right])] += 1
7        if right - left + 1 >= k:
8            odd_freq = [f for f in freq if f % 2 == 1]
9            even_freq = [f for f in freq if f % 2 == 0 and f > 0]
10            if odd_freq and even_freq:
11                max_odd = max(odd_freq)
12                min_even = min(even_freq)
13                max_diff = max(max_diff, max_odd - min_even)
14            freq[int(s[left])] -= 1
15            left += 1
16    return max_diff if max_diff != float('-inf') else -1

ℹ

Complexity note: The time complexity is O(n) because we traverse the string once with the sliding window approach, and the frequency calculations are constant time due to the fixed number of characters (0-4).

1Understanding character frequency is crucial for solving this problem.
2The sliding window technique allows for efficient substring analysis without generating all possible substrings.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.