Maximum Difference Between Even and Odd Frequency II

Hard

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StringsSliding Windows

You are given a string s and an integer k. Your task is to find the maximum difference between the frequency of two characters, freq[a] - freq[b], in a substring subs of s, such that:

subs has a size of at least k.
Character a has an odd frequency in subs.
Character b has a non-zero even frequency in subs.

Return the maximum difference.

Note that subs can contain more than 2 distinct characters.

Example 1:

Input: s = "12233", k = 4

Output: -1

Explanation:

For the substring "12233", the frequency of '1' is 1 and the frequency of '3' is 2. The difference is 1 - 2 = -1.

Example 2:

Input: s = "1122211", k = 3

Output: 1

Explanation:

For the substring "11222", the frequency of '2' is 3 and the frequency of '1' is 2. The difference is 3 - 2 = 1.

Example 3:

Input: s = "110", k = 3

Output: -1

Constraints:

3 <= s.length <= 3 * 10⁴
s consists only of digits '0' to '4'.
The input is generated that at least one substring has a character with an even frequency and a character with an odd frequency.
1 <= k <= s.length

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the input array 'nums'?
Can the elements in the array 'nums' be negative, zero, or non-integer?
If all subarrays have a difference of zero or negative, what value should I return?
Are we looking for contiguous subarrays only?
Are there any specific considerations for handling very large differences in frequency between even and odd numbers that could lead to integer overflow?

Brute Force Solution

Approach

The brute force method for this problem means checking every single possible group of numbers within the given set. We calculate a 'score' for each group based on how often even and odd numbers appear, and then pick the group with the best score. It's like trying out every possible combination.

Here's how the algorithm would work step-by-step:

Consider every possible starting point within the set of numbers.
For each starting point, consider every possible ending point that comes after it or is the same as it.
For each combination of starting and ending points, we now have a group of numbers.
Count how many times each number appears an even number of times, and count how many times each number appears an odd number of times within that group.
Calculate the 'score' of that group by subtracting the number of odd-frequency numbers from the number of even-frequency numbers.
Compare the current group's score to the best score we've seen so far. If it's better, remember this score and the group that created it.
Once we've checked every possible group of numbers, the best score we remembered is the answer.

Code Implementation

def max_difference_even_odd_frequency_brute_force(numbers):
    max_difference = float('-inf')

    for start_index in range(len(numbers)):
        for end_index in range(start_index, len(numbers)):
            sub_array = numbers[start_index:end_index + 1]
            frequency_map = {}

            for number in sub_array:
                if number in frequency_map:
                    frequency_map[number] += 1
                else:
                    frequency_map[number] = 1

            even_frequency_count = 0
            odd_frequency_count = 0

            # Count numbers with even or odd frequencies
            for number, frequency in frequency_map.items():
                if frequency % 2 == 0:
                    even_frequency_count += 1
                else:
                    odd_frequency_count += 1

            # Calculate the difference for this sub-array
            difference = even_frequency_count - odd_frequency_count

            # Update max_difference if necessary
            if difference > max_difference:
                max_difference = difference

    return max_difference

Big(O) Analysis

Time Complexity

O(n^3) – The brute force approach involves iterating through all possible subarrays. There are two nested loops to define the start and end indices of each subarray, resulting in O(n^2) subarrays. For each subarray, we count the frequency of each number, which takes O(n) time in the worst case (where all elements are unique). Therefore, the overall time complexity is O(n^2 * n) which simplifies to O(n^3).

Space Complexity

O(N) – The brute force approach calculates frequencies of numbers within each sub-array using a hash map (or similar data structure) to store counts. In the worst case, all N elements of the input array could be distinct within a particular sub-array, requiring O(N) space for the frequency counts. While individual sub-array calculations reuse the same map, each of the O(N^2) sub-arrays effectively needs to track its own frequency information up to size N. The other variables used (start index, end index, max score) use constant space, but the frequency map dominates space usage, so the space complexity is O(N).

Optimal Solution

Approach

The goal is to find the largest difference between how often even numbers appear and how often odd numbers appear in a set of numbers. The key idea is to track the running total of this difference and reset it whenever it becomes negative, as a negative difference only hurts our overall maximum.

Here's how the algorithm would work step-by-step:

Start by assuming the biggest difference we've seen so far is zero.
Go through each number in the set one by one.
If the number is even, add one to a running total; if it's odd, subtract one.
If the running total ever goes below zero, reset it back to zero. This is because a negative running total cannot contribute to a larger difference later on.
After each number, check if the running total is bigger than the biggest difference we've seen so far. If it is, update the biggest difference.
The biggest difference at the end is the largest possible difference between even and odd number frequencies you can achieve.

Code Implementation

def find_maximum_difference(numbers):
    maximum_difference = 0
    running_difference = 0

    for number in numbers:
        # Increment if even, decrement if odd.
        if number % 2 == 0:
            running_difference += 1
        else:
            running_difference -= 1

        # If running total is negative, reset it.
        if running_difference < 0:
            running_difference = 0

        # Update the maximum difference if needed.
        if running_difference > maximum_difference:
            maximum_difference = running_difference

    return maximum_difference

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n exactly once. Inside the loop, it performs a constant number of operations: checking if a number is even or odd, updating the running total, and comparing the running total to the maximum difference seen so far. Since the number of operations within the loop doesn't depend on the size of the input, the overall time complexity is directly proportional to n.

Space Complexity

O(1) – The algorithm uses a single variable to keep track of the running total and another to store the maximum difference seen so far. These variables consume a constant amount of space regardless of the number of elements, N, in the input set. Since no auxiliary data structures that scale with the input size are used, the space complexity remains constant. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array	Return 0, as there's no subarray to consider.
Array with a single element	Check if the single element is even or odd; return 1 if even, -1 if odd.
Array with all even numbers	The maximum difference will be the length of the array, representing the subarray containing all even numbers.
Array with all odd numbers	The maximum difference will be the negative of the length of the array, representing the subarray containing all odd numbers.
Array with alternating even and odd numbers	Kadane's algorithm correctly computes the maximum difference in this mixed scenario.
Large input array size (performance)	Using Kadane's algorithm or a similar O(n) approach ensures efficient scaling to large inputs.
Array contains both positive and negative numbers but treats them as parity i.e., even/odd	The algorithm should function correctly with both positive and negative numbers because even/odd check works fine for negative numbers too; negative even numbers will not cause any problems.
Integer overflow when calculating frequency differences with extremely long arrays	Use a data type that can accommodate larger numbers (e.g., long) to prevent potential overflow when calculating frequency differences for very large arrays.

Empty input array

How to Handle:

Return 0, as there's no subarray to consider.

Array with a single element

How to Handle:

Check if the single element is even or odd; return 1 if even, -1 if odd.

Array with all even numbers

How to Handle:

The maximum difference will be the length of the array, representing the subarray containing all even numbers.

Array with all odd numbers

How to Handle:

The maximum difference will be the negative of the length of the array, representing the subarray containing all odd numbers.

Array with alternating even and odd numbers

How to Handle:

Kadane's algorithm correctly computes the maximum difference in this mixed scenario.

Large input array size (performance)

How to Handle:

Using Kadane's algorithm or a similar O(n) approach ensures efficient scaling to large inputs.

Array contains both positive and negative numbers but treats them as parity i.e., even/odd

How to Handle:

The algorithm should function correctly with both positive and negative numbers because even/odd check works fine for negative numbers too; negative even numbers will not cause any problems.

Integer overflow when calculating frequency differences with extremely long arrays

How to Handle:

Use a data type that can accommodate larger numbers (e.g., long) to prevent potential overflow when calculating frequency differences for very large arrays.